Vedic Mathematics(orignal Book) 1k1543

, 8 q -a t c -v , ! n w.r. 1 , f * etc. case 0 1 7), we may take —3 as the common geometrical ratio and will find the Geometrical Progression intact; and naturally the product of each 3 Q ar -s e -• - 1 a . , ,. ~“ T T 2 > b > 1 remamder-digit by the 0 7 6 9 2 3 last digit remains in tact too and gives us the same answer: *076923! We now on to still another and easier method which comes under the EkddJdka Sutra which we have expounded and explained at sufficient length already (in the first chapter) and which therefore we need only summarise and supplement here but need not elaborate again. The Ekddhika Sutra (which means ‘by the preceding one increased by one’ ) has already been shown at work in a number of ways and in a number of directions and on a number of occasions and will similarly come into operation still further, in many more ways and in many more contexts.

( 204 ) Numbers ending ■in Nine (i) If and when the last digit of the denominator is 9, we know beforehand that the equivalent recurring decimal ends in 1. (ii) In the case of i V the last but one digit is I; we increase it by 1 and make it 2. In we work with 2 + 1 = 3 . In and in we operate with 4 and 5 respectively and so on. (iii) In the multiplication-process (by Ekadhika Purva), in all these cases, we put 1 down as the last digit (i.e. in the right-hand-most place); and we go on multiplying that last digit (1) from the right towards the left by 2, 3, 4 and 5 respec tively ; and when thete is more than one digit in th&t product, we set the last of those digits down there and carry the rest of it over to the next immediately preceding digit towards the left. (iv) When we get D <*N as the product, we know we have done half the w ork; we stop the multiplication there; and we mechanically put down the remaining half of the answer (by merely taking down the complements from NINE). (v) The division-process (by Ekadhika Sutra) follows the same rules (vide Supra). ' ~ (1) We may first consider the fraction as our first illus tration of the method described : (i) Putting 1 as* the last digit and continually multiplying by 2 towards the left, we get the last four digits (towards the left) without the least difficulty.

............................................... . 8 4 2 1 9 4 7 3 6 (ii) 8 x 2 = 1 6 . Therefore put 6 down immediately to the left of 8 (with 1 to carry over). 6 x 2 + th e 1 carried over—13. Put the 3 to the left of the 6 (with 1 to carry over). 3 X 2+1 = 7 Set it down before the 3 (with nothing to carry over). 7 X 2 = 14. Therefore put the 4 before the 7 (with 1 to carry over) 4 x 2 + th e 1 carried over=9. (iii) We have thus got 9 digits by continual multiplication from the right towards the left. And now 9 x 2 = 1 8 (which is D * N). This means that half the work is over and that the earlier 9 digits are obtainable by putting down the complements

( 206 ) (from NINE) of the digits already determined.

So, we have

TV = 0 0 5 2 6 3 1 5 7 8/9 4 7 3 6 8 4 2 i (2) Let us now examine the case of *VBegin with 1 (as usual) at the extreme right end and go on multiplying by 3 each time, “ carrying over” the surplus digit or digits (if any) to the left (i.e. to be added to the next product to be determined. Thus, when we have obtained 14 digits i.e. .......................... 9 6 5 5 1 7 2 4 1 3 7 9 3 1, we find that we have reached 28 ; we know we have done half the w ork; and we get the first 14 digits by simply subtracting each of the above digits from NINE. •0 3 4 4 8 2 7 5 8 6 2 0 6 8/ *l* = 0 -6 3 4 4 8 2 7 5 8 6 2 0 6 8/96 5 5 1 7 2 4 1 3 7 9 3 i (3) Next let us take ^ Take 1 again at the extreme right end and continually multiply by 4 from the right to the left. Thus, we have : *y=-02564i. Note in this case that V 39 is a multiple of 3 and 13 and not a prime number (like 19 and 29) and v 3 and 13 give only 1 and 6 recurring decimals, there is a difference in its behaviour i.e. that the two halves are not complementary with regard to 9 but only in relation to 6 ! In fact, D ^ N (i.e. 38) does not come up at all as an interim product (as 18 and 28 did). And so, the question of complements from 9 does not arise at a ll; and the decimal equivalent has only 6 figures (and not 38) ! The reason for this is very simple. As as fa has only 6 recurring decimals in its decimal equivalent and because, for reasons to be explained a little later, this decimal equivalent of T* 3 is exactly divisible by 9, much more therefore is it divisible by 3. And, consequently, when we divide it by 3 and exhaust the six digits, we find that there is no remainder left. In other words, fa has only 6 digits in its recurring decimal shape. These have been obtained by the self-same Ekadhika pro cess as served our purpose in the case of fa and fa

( 206 ) We next take up and examine the case of •which, besides following the rules hereinabove explained, has the additional merit of giving us the clue to a still easier process for the con version of vulgar fractions into their recurring decimal shape: ( !) _ * _ _ 1 _ * *142867 * * 7X 7 7 (i) If we go on dividing 1 by 49 or *142857 by 7 (until the decimal begins to recur), we shall doubtless get our answer. But this will mean 42 steps of labo rious working and is therefore undesirable. (ii) We therefore adopt either of the Ekddhika methods and go on multiplying from right to left by 5 or dividing from left to right by 5. (iii) On completing 21 digits, we find 48 (i.e. D N) coming up and standing up before u s ; and we mechanically put down the other 21 digits as usual (by the subtrac tion, from 9, of the digits already obtained). And the answer i s :

(iv) And this gives us the clue just above referred to about a still easier method (than even the Ekddhika ones) for the con version of vulgar fractions into recurring decimals. And it is as follows: By actual division (of 1 by 49), we observe that the successive remainders are in Geometrical Progression (with the common ratio 1 : 2) that the dividends are similarly related and that each set of two digits in the quotient is also so related to its predecessor.

49)1 *00 ( *20408 98 200 196 400

In other words, this connotes and implies that,

after putting down 02, we can automatically put down 04, 08* 16 and 32 and so on.

( 907 ) But when we reach 64, we find that 2 X 6 4 »1 2 8 i.e. has 3 digits. All that we have to do then is to add the 1 of the 128 over to the 64 already there, turn it into 65 and then put down not 28 but the remaining part of double the corrected figure 65 (i.e. 30) and carry the process carefully on to the very end (i.e. until the decimal starts to recur). We therefore have : 0* 0 2 0 4 0 8 1 6 3 2 6 5 3 0 6JL 2 2 4 4 8

l

9 1 9 5 9 1 8 3 6 7 3^4 6 9 3 8J7 7j5 5 i 1 This new method does not apply to ^11 cases but only to some special cases where the Denominator of the given vulgar fraction (or an integral multiple thereof) is very near a power of ten and thus lends itself to this kind of treatment. In such cases, however, it is the best procedure of all. N ote:— The rule of complements (from 9) is actually at work in this case too ; but, inasmuch as (for reasons to be explained hereafter), the actual total number of digits is 42, the first half of it ends with the 21st digit and as we have been taking up a group of two digits at each step, we naturally by- the 21st digit (which is concealed, so to speak, in the middle of the 11th group). But, even then, the double-digit process is so very simple that continuation thereof can present no difficulty. Other Endings So far, we have considered only vulgar fractions whose denominators end in 9. Let us now go on to and study the cases of fa, fa , fa and other such fractions (whose denomina tors end not in 9 but in 1, 3 or 7). (i) Here too, we first make up our minds, at sight, as regards the last digit of the decimal equivalent. Thus, Denominators ending in 7, 3 and 1 must neces sarily yield decimals ending in 7, 3 and 9 (so that the product of the last digit of the denominator and the last digit o f the decimal equivalent may end in 9.) Let us start with the case of

(

208

)

(ii) Put down \ in the shape (iii) Take 5 (one more than 4) as the Ekadhika Purva for the required multiplication ox division (as the case may be). (iv) Thus start with 7 at the right end

85 7 23

(v) Multiply it by 5 and set down 35 as shown in the marginal chart. (vi) Multiply 5 by 5, add the 3 to set 28 down in the same way. Now, 5 X 8 + 2 = 4 2 .

the product and

But that is D

N

142

57

Therefore put 142 down as the first half (according to the complements rule) ^=-142/857 Or 1 =

'

The Ekddhika being 5, divide 7 by 5 1 4 2 /8 5 7 and continue the division as usual 2 1 4 / (with the same rule of procedure). After getting the three quotient-digits 1, 4 and 2 you find 42 as the remainder before you. So tackle the last 3 digits (according to the complements rule) and s a y :

(2) Let us now take the case of Tfr — (i) The last digit is 3.*.the last digit (in the answer) will be 3.

*0 7 6 / 9 2 3 •07 6 / 9 2 3 2 3

(ii) The Ekddhika (multiplier or divisor) is .*.4. (iii)

ts = sV After 3 digits (whether by multipliction or by division), 36 (Dio N) comes up. So, the other half is mechanically set down. And we say :

T\ = ‘076/923

( *#

)

(3) Next, let us take (i) The last digit is 1. The last digit (of the answer) will be 9.

0/ 9 9j

(ii) The Ekadhika (in both ways) is 10. (iii) Immediately after the very first digit, we get 90 (which is 9 0 M9) before us. So, the complements* rule operates. (iv) And, in either case, we get Tix= ,0i/9 (4) * * = (giving 7 as Ekadhika) and 3 as the last digi£ of the answer. (By both methods, Multiplication and Division). 1 = 0 *0 4 3 4 7 8 2 0 0 8 0 ) ” 9565 2 173913 > (5) T1T= Tj v (giving 12 as Ekadhika and 7 as the last digit) By both the methods (multiplication and division), we have: ^T= Tf * —'0 5 8 8 2 3 5 2 / 9 4 1 1 7 0 4 7 The Code Language at Work. Not only do the Vedic Sutras tell us how to do all this by easy and rapid processes of mental arithmetic; but they have also tabulated the results in the shape of special sub-Sutras (containing merely illustrative specimens with a master-key for “ unlocking other portals” too). The abstr details (and the master-key) are not given here ; but a few sample-specimens are given of the way in which the code and the Ekanyuna SHtra (explained in Chapter 2) can be utilised for the purpose of postulating mental one-line answers to the questions in question. The three samples read as follows: (1) ^TT^; (Kevalaih Saptakarp, Gunydt) (2) : (Kalau Ksudrasasaih): and (3) *1% (Kamse Ksamaddha-Jchalairnialaih) In the first of these, Saptaka means ‘seven’ ; and Kevalaili represents 143 ; and we are fold that, in the case of seven, our multiplicand should be 143 ! 27

( 210 ) In the second, Kalau means 13 and Ksudmsasaih repre sents 077 ; and we are told that the multiplicand should be 077 ! and, In the third, Kamse means 17 ; and Ksdma-ddha-khalairmalaih means 05882353 ; and we are told that the multiplicand should be this number of 8 digits ! Now, if we advert to the “ Ekanyuna,” corollary of the Nikhilam chapter (on multiplication), we shall be able to remind oursleves of the operation in question and the result to be achieved thereby. Let us do the multiplications accord ingly (as directed) and see what happens. (1) In the case of 7 (as denominator), 143X999=142/857 ; and lo ! these are the six recurring decimal digits in the answer ! (2) In the case of 13, 077 X 999=076/923 ; and these are the six digits in the recurring decimal equivalent of TS • and (3) In the case of 17, 05882353 X 99999999=05882352/ 94117647 ; and these are the 16'recurring digits in the recur ring decimal equivalent of ! In all the 3 cases we observe the Rule of Complements (from 9) at work. And the sub-Sutra merely gives us the necessary clue to the first half of the decimal and also a simple device (Ekanyunena) for arriving at the whole answer ! And all this is achieved with the help of the easy alphabet-code ! These results may therefore be formulated as follows: i _143x999__142857_ . \4.9^ j . T 9 9 9 9 9 9 9 9 9 9 9 9

t

q72 x 9 9 9 = 076923==.Q769 2 3 ; and

TS

999999

999999

! _ 0 5 8 8 2 3 5 3 X 99999999 lT

9999999999999999 =

-6 5 8 8 2 3 5 2 /9 4 1 1 7 6 4 7 * !

And, by CROSS-multiplication, we get from the above, the following results: (1 ) 7 X 1 4 2 8 5 7

=999999;

(2 ) 1 3 X 0 7 6 9 2 3 = 9 9 9 9 9 9 ; an d (3 ) 1 7 X 0 5 8 8 2 3 5 2 /9 4 1 1 7 6 4 7 = 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 (1 6 d ig its in a l l ) !

( 211 ) And, just in ing, we may note that this is the reason why, in the case of all these vulgar fractions, the last digit of the denominator (9, 3, 7 or 1, as the case may be) gives, 1,3,7 or 9 before-hand as the last digit of the equivalent recurring decimal fraction ! The Remainder-Quotient Complements-Cyeles. We have already - again and again - noted the fact that, in the various typical cases observed and analysed by us, the two halves of the quotients (added together) give us a series of NINES. We shall now proceed a little bit further and try to see if there be any such (or similar) rule governing the remainders. For the purpose of the necessary experimentation and investi gation, let us take up a more detailed consideration of the remainders obtained in each case by successive divisions of the numerator by the denominator; and let us start with We know the successive Remainders are 3, 2 , 6 , 4, 5 and . We know already that, on reaching 6 (i.e. D ^ N ) as the remainder, half the work has been completed and that the complementary half is about to begin. Putting the above 6 figures, therefore, into two rows of three ^ $ j figures each, we have : and we observe that each vertical column (of one digit from the 1

upper row and of one from the lower one) gives us the same total (i.e. 7 )!

4

5

1

^— j

N.B. :—As our divisor is 7, it is but natural that no remainder higher than 6 is permissible i.e. that the only possible remainders are 1 , 2 , 3, 4, 5 and 6 . (And these are the ones we actually find). Let us now take up the case of and note what happens. The successive remainders are 1 0 , 9, 1 2 , 3, 4 and 1 (the highest of which is 1 2 ). And when they are placed 109 1 2 / 3 4 1 in two rows, we find here too, that the last 10 9 12 three remainders are complements— from 3 4 ^ 13—‘of the first three remainders.

13 13 \ 13

( 212 ) In the case of fa the successive remainders are i 10 15 14 4 6 9 5 16/7 2 3 13 11 8 12 1 7 2 3 13 11 8 12 1 17

17

17

17

17

17

17

17

The last 8 remainders are thus complements—from 17—of the first eight ones ! \n the case of 19, the Remainders a re: 10 15 12 6 3 11 15 17 9 4 7 13 16 8 4 2 19 19 19 19 19 19 19 19

18 1 19

Here again the first nine remainders,

when added

successively to the next nine, give 19 each time. Thus, it is clear that, whereas the quotient-halves are uniformly complements from nine, the remainder-halves are complements from the individual divisor in each case. And this further reduces our labour in making out a list of the Re mainders. Multiples of the Basic Fractions. Thus far, we have dealt with vulgar fractions whose numerator is unity. But what about fractions which have some other Numerator ? And the answer is : “ There are several simple and easy methods by which, with a tabulated list before us of the results obtained by one or more of the processes above expounded, or even independently, we can readily put down the Recurring Decimal equivalents of the vulgar fractions of the type just under discussion. Let us, as usual, start with \ and frame a chart as follows * 1 — 142857 T — 2 ___ 285714 r 3_. 42857i T— * = 571428 5 __ T— 714285*; and T = 857142

( 21$ ) In this chart, we observe that, in all the *‘proper” fractions having 7 as their Denominator, (i) The same six digits are found as in the case of ^ ; (ii) they come up in the same sequence and in the same direction as in the case of \ ; (iii) they, however, start from a different starting-point but travel in “ Cyclic” order (in what is well-known as the “ Clock-wise” order). (iv) and with the aid of these rules, one can very easily obtain the recurring decimal equivalent of a vulgar fraction whose numerator is higher than 1. In fact, a person who is actually looking at a statement (on a board, a piece of paper, a slate etc.,) to the effect that ^=.142857, has several easy alternative processes to choose from, for determining the decimal equivalents of all the other five possible fractions having the same denominator (i.e. 7). They are as follows: The First Method. 1. The verious digits can be numbered and marked in ascending order of magnitude, thus : (i) Unity being the least of (1) (3) (2) (6) (4) (5) them, the cycle for \ starts 1 4 2 8 5 7 with one as its starting point, travels in clock-wise cyclic order and reads: i 42857 ; (ii) 2 being the second, f starts with 2 and gives us the answer *285714; (iii) There being no 3 at all, the third digit in ascending order is 4. So f begins from 4 and reads : *42857i ; (iv) The next digit (i.e. the 4th) in ascending order actually being 5, 4/7 begins with 5 and gives : •571428;

( m

)

(v) The fifth digit actually being 7, 5/7 commences with 7 and reads : *571428; and (vi) The 6th and last digit being 8, the sixth and last fraction (i.e. f) starts with 8 and says: *857142 ThiS* is the first method. Yes, but what about those cases in which the number of decimal places is more than 10 and thus, in the tabulated answer before us, some digits are found more than once ? Yes, it is perfectly true that, just as some digits are found absent [as in the case of \ (just seen)], there are other cases where the same digits are found more than once. In fact, in every case wherein the number of decimal places is more than 10, this is bound to happen ; and provision too there must be against it. In fact, the remedy is very simple i.e. that, even where digits occur more than once, there still are gradations; and, if these are taken into , the cyclic order and the ascending order of magnitude will still operate and serve their purpose. For example, in the case of ^T, we have .05882.......at the very commencement; and there are two eights before us. Yes, but 88 is greater than 82; and therefore we should take 82 first and 88 afterwards and do our numbering accordingly : T*T= *6 5 8 8 2 3 5 2 9 4 1 1 7 6 4 7 (1) (10) (15) (14) (4) (6) (9) (5) (16) (7) (2) (3) (13) (11) (8) (12) Thus, Xx7 starts with zero; X2T with 11; TST ; with 17 ; with 23 ; t57 with 29; X6T with 3 ; TT7 with 41; X8T with 47; T®7 with 52; with 58 ; ^ with 6; with 70; with 76; with 82 ; with 8 8 ; and TT with 9 . The arranging in ascendmg order of magni tude has, of course, to be done carefully and correctly. But it must be itted that, although the procedure of counting and numbering is quite reasonable and scientifically correct, yet it is rather cumbrous, clumsy and tiring. Hence the need for other methods.

( 2U> ) Yes ; but what again about the cases wherein the number of digits in the decimal equivalent is much less than the deno minator of the vulgar fraction in question and has thus no scope for meeting all the possible demands ? Yes, fa is such a case. The number of possible multiples is 12; and the number of digits in the decimal equivalent is only 6. (for -076/923). What is the remedy? The remedial provision is that a multiple or two will do the trick quite satisfactorily and neatly. Now, *076923* /. (By simple multiplication by 2), ys = *i 53846 And now, there are twelve digits in all; and these can meet the needs of all the possible multiples. Thus— TV = *076923 ; and -

*153846

t s = ’ 230769 ; and x4* = *307692 •384615 ; and x6* = *461538 •538461 ; and x8* = * 615384 •692307 ; and ^ = *7 6 9 2 3 0 = •846153 ; and

*923076

The procedure is there and is quite correct. But, after all, on# must confess that, even with this device, this counting and numbering procedure is still a cumbrous, clumsy and tiring process. Hence, let us repeat, the need for other methods. The Second Method The second method is one wherein we avoid even this num bering and marking etc., and (in accordance with the Adyam Adyena rule), multiply the opening digit or digits of the basic decimal fraction (*142857) and determine, therefrom the starting point for the multiple in question. Thus, *142857 \ starts with *14.......... \ f should start with *28 etc., and (in clockwise cyclic order) give *285714 ; f ought to start with *42 etc., and give *428571 ;

( sie ) £ must start with *56; but there is no *56 but only *57 before u s ; and so (making allowance for a possible nay, the actual— carrying over of a surplus digit from the right leftward), we start with 57 and say: $=*571428; Similarly f should start with *70 ; but (for the same reason as in the immediately preceding case), it actually starts with •71 and gi^es us : *714285; and $ should start with *84, actually starts with *85 and yields the answer : 857142! This is the Second method. The Third Method The third process is very similar; but it bases itself not on Adyam Adyena but on Antyam Antyena. In other words, it deals not with the opening digit but with the closing one. Thus,

and

•V \ ends with 7, .% $ must end with 4 / . $ should end with 1 $ ought to end with 8 £ should end with 5 $ must end with 2 ,\

It It It It It

is is is is is

*i 42857 *285714 -428571 *571428 *714285 *857142

This is the third method and the easiest and therefore the best of the lot. Independent method The above described methods are all for the utilisation of our knowledge of the decimal shape of a fraction whose nume rator is unity, for deriving the corresponding decimal form of any multiple of that fraction. This is all right, so far as it goes. But what about a person who ha$ not got such a readyto-hand table to refer to ? In such a case, should one newly prepare the basic chart and then manipulate it—cyclically— (in one of the ways just explained), for getting the required result ? That would, of course, be absurd. For use by such persons, we have too, a totally independent method, by which,

< 217 ) ■without resorting to any such previously prepared (or newly prepared) table, one can readily deal with the particular fraction on hand! And the whole modus operandi is exactly the same as has been already explained in respect of the basic fraction and without the slightest difference or deviation in any particular whatsoever therefrom. For example, suppose you have to decimalise f. Your last digit will be 1. and as f= f& , your Ekadhika Purva will be 5. Now, go on dividing by 5, in the usual manner; and you get the chart, as explained in the margin: f = H = * 4 2 8/571 1 4 2/ After you get the lirst three digits 4, 2 and 8, you find that your dividend is 2 8 ; but this is D « N (i.e. 49—21). So you may stop here and put the last three quotient-digits down as 5, 7 and 1 (the complements, from nine, of the digits already found). Or you may continue the division till you get 21 as the dividend; and as this was your starting-point, you may put the 6 digits down as a “Recurring11’ decimal. Thus | = *428571 Try this with r%, tt> t v an(i so on> with any number o f cases. And you will always find the same thing happening right through all o f them. Thus, for those who do not have a tabu lated schedule before them, this absolutely independent method is also there: and you can make full use of it. N ote: —1,

In this independent method, it should also be noted that if we have to decimalise f , f , f , f , etc., we have merely to divide 10, 20, 30, 40, 50 etc., by 7 and put down that remainder as the first remainder in each particular case and that the work can be done automatically thereafter.

2.

or, we may pre-decide the last digit in each case by taking the last digits of 7, (1) 4, (2) 1, (2) 8, 3(5), (4) 2 as the last digits of the decimal equivalent of b b

28

f and f !

( 218 ) Recapitulation (and Supplementation) Over and above the ones expounded and explained herein above, there are several other very instructive and interesting principles, features and characteristics characterising this question of the conversion of vulgar fractions into decimal ones (in respect of the remainders, the quotients etc.). For the benefit of the# students, we propose now to recapitulate, summarise, supplement and conclude this portion of the subject: (1) As regards the remainders, we have noted that, as soon as D ^ N comes up before us as a remainder, the remaining remainders are all complements—from the divisor (i.e. the denominator)—of the remainders already obtained ; (2) This automatically means that the quotient—digits already obtained and the quotient—digits still to be found, are complements from Nine ! (3) If we take any remainder and multiply it by the

Garamanka (the last digit), the last digit of the product is actually the quotient at that step.

(The formula here is wftr

(Sesani Ankena Caramena) which is therefore of the utmost significance and practical utility in mathematical computations. For instance, (1) \ The remainders are 3, 2, 6, 4, 5 and 1. Multiplied by 7 (the Garamanka) these remainders give successively 21, 14, 42, 28, 35 and 7. Ignoring the left-hand side digits, we simply put down the last dight (Charamanka) of each product; and lo ! We get ^=*142857 ! (2) fa The remainders are 10, 9,12, 3, 4 and 1. Multiplied successively by 3 (the last digit), these remainders give 30, 27, 36, 9, 12 and 3. Ignoring the previous digits, we write down merely the Caramdnkas the last digit) of each product; and lo! f a = *079623! (3) fa The remainders are 10, 15, 14, 4, 6, 9, 5, 16/7, 2, 3, 13, 11, 8, 12 and 1. Multiplied by 7, they give us successively:

( 210 ) 70, 105, 98, 28, 42, 63, 35, 112, 49, 14, 21, 91, 77 56,84 and 7. Dropping the surplus (i.e. left-side) digits and putting down only the Carmankas (the right-hand most digits), we have T»T= -05882352/94117647 In fact, the position is so simple and clear that we need not multiply the whole digit, write down the product and then drop the surplus digit (or digits). We need only put down the Charamanka (the right-hand-most digit) at the very outset (as each step) and be done with i t ! (4) The Geometrical-Progression =■ character of Remainders gives us a clue to the internal relationship between each remainder (and its successor) (or its predecessor)! Thus, as we know one remainder, we practically know all the rest of them Thus, In the case of ^ : As we know the first remainder is 3, we can multiply any remainder by 3, cast out the sevens (if any) and immediately spot out and announce the next Remainder. 3 X 3 = 9 ; 9—7 = 2 ; ,\ 2 is the second remainder 2 x 3 = 6 . This is the third rdmainder. As 6 is- D w N, we may stop here and (by the rule of com plements from the denominator), we may put down 4, 5 and 1 as the remaining three remainders. Or, if we overlook the D » N rule or prefer to go on with our multiplication by 3 (the Geometrical ratio), we g e t : 6 x 3 = 1 8 ; 18—14=4 ; and this is the 4th remainder. 4 X 3 = 1 2 ; 12—7 = 5 ; and this is the 5th remainder. 5 X 3 = 1 5 ; 15—1 4 = 1 ; and this is the 6th (and last) Remainder. We have thus obtained from the first remainder, all the remainders: 3, 2, 6, 4, 5 and 1. And from these, by multiplication by the Carama'hka (7), we get all the 6 quotient-digits (as explained above): •1, 4, 2, 8, 5 and 7

the

( 220 ) This is not all. Instead of using the first remainder (3) as our Geometrical Ratio, we may take the second one (2), multiply each preceding group of 2 remainders by 2 and get 32, 64 and 51 (for, by casting out the sevens, 6 x 2 —7 = 5 ; and 4 X 2 —7=1). And multiplying these 6 digits by 7, we again get the Caramankas 1 4 2 8 5 7 (as before). Or we may take help from the third remainder (i.e. 6), multiply the preceding group of 3 remainders and get 3 2 6, 4 5 1 (for, by casting out the sevens, 3 x 6 —14=4 ; 2 x 6 —7 = 5 ; and 6 X 6 —35=1. And, multiplying these (same) 6 digits by 7, we again obtain the Caramankas 142857 as before. This procedure is, of course, equally applicable to the fourth and fifth remainders (i.e. 4 and 5) and can get us the same result as before This is doubtless purely academical and o f no practical utility. But we are discussing a principle, nay a universally operating mathematical law and must therefore demonstrate its actual universality of application. So, if we take the 4th remainder (i.e. 4) and multiply the preceding group of four remainders by 4, we again get 3264/51 (For, 4 x 3 - 7 = 5 ; 4 x 2 - 7 = l ; 4 x 6 - 2 1 = 3 ; 4 x 4 - 1 4 = 2 ) ; and the only difference is that the first two digits are found to have already started repeating themselves! If we now take the 5th remainder (i.e. 5) and multiply the preceding group of 5 remainders by 5, we again get 32645/1... (for 3 X 5 —1 4 = 1 ; 2 X 5 —7 = 3 ; 5 x 6 —2 8 = 2 ; 5 x 4 - 1 4 = 6 ; 5 X 5 -2 1 = 4 ; And, if we follow the same procedure with the 6th remainder (i.e. 1) and multiply the group of preceding remainders by 1, we will, of course, get th^ same preceding remainders over again! (5) In the case of 17, the first four remainders are : 10, 15, 14 and 4. As 4 is a manageable multiplier, we may make use of it as a convenient and suitable remainder for this purpose. Let us therefore multiply the group of four Remainders (already

( 221 ) found) by 4 and cast out the seventeens (wherever necessary). And then we find: 4 X 1 0 -3 4 = 6 10, 15, 14,4 4 x 1 5 —51=9 10, 15, 14, 4, 6 4 x 1 4 —51=5 10, 15, 14, 4/6, 9, 5, 16/ 4 X 4 = 1 6 . But as D
xv :

•05 8 8 2 3 5 2 /9 4 1 1 7 6 4 7 Still another method

Besides (1) the corollary-Sutra (2) each remainder X the last digit method, (3) the Ekadhika process from right to left and (4) the Ekadhika method from left to right, there is still another method whereby we can utilise the Geometrical-Progre ssion relationship and deduce the same result by a simple and easy process. And it is this, namely, that as soon as we come across a clear ratio between one remainder (or dividend) and another, we can take that ratio for granted (as being of universal application) and work it out all through. For example, In the case of 19, we have 10 and 5 as the first two remainders and we note that 5 is just one-half of ten. Keeping this ratio in view, we can deduce that the next remainder should be one-half of 5. But, as 5 is not exactly divisible by 2, we add 19 to it, make it 24 and put down its half (i.e. 12) as the next remainder. The 12 10, 5, 12, 6, 3, 11, 15, 17, 18 gives 6, 6 gives 3, 9, 14, 7, 13, 16, 8, 4, 2, 1 --------------------------------- -— ------- ------3(+19) gives 11, 11(— (-19) gives 15, 1 5 (+ 1 9 ) gives 17 & (1 7 + 1 9 ) gives 18. And we stop there and put down the remaining half of the remainders by subtractions from 19. Having thus got the remainders, we multiply the Caramdnkas by 1 (the last digit of the answer) and we get the quotient-digits automatically.

1

( m N .B .

)

The ratio in question may be noticed at any stage of the work and made use of at any point thereof.

In the case of Xx7, we have the remainders 10 and 15 at the very start. We can make use of this ratio immediately and throughout, with the proviso that, if and when a fractional product is threatened, we can take the denominator (or as many multiples thereof as may be necessary) for making the digit on hand exactly divisible by the divisor on hand. I Thus, in the case of ^7, we have the remainders 10 and 15 to start with (the ratio being 1 to 1J). So, whenever one odd number crops up, its successor will be fractional. And we get over this difficulty in the way just explained. And when we get a remainder which is numerically greater than the divisor, we cast off the divisor and put down the remain der. Thus, 10 gives us 15; 15 (+ 17) gives 10,15,14, 4, 6 ,9 , 5,16 us 48 i.e. 14; 14 gives us 21 i.e. 4; 7, 2, 3, 13, 11, 8, 12, 1 4 gives us 6 ; 6 gives us 9 ; ------------------------------------- — 9(+17) gives us 39 i.e. 5; (5+17) gives us 33 i.e. 16. And there we can stop. Number of Decimal Places Students generally feel puzzled and non-plussed as to how to know beforehand the number of decimal places which, on division, the decimal equivalent of a given vulgar fraction will actually consist of. In answer hereto, we must point out that, having—in the immediately preceding sub-section on this subject—made a detailed, analytical study of the successive remainders, we have, in every case before us, practically a tabulated statement from which (without actual division to the very end) we can postulate beforehand all the forthcoming remainders. And the tabulated statement has the further merit that it can be prepared, at any time, at a moment’s notice I

( 223 ) All this means, in effect, that, (i) As soon as 1 (or other starting point) is reached (in our mental analysis), we will have completed the whole work of'decimalisation and therefore know the actual number of decimal places coming ahead. The cases b tV> etc-> have a11 Proved this(ii) As soon as we reach the difference between the numerator and denominator, we know we have done half the work and that the other half is yet to come. The cases of \ etc., (which we have dealt with in extenso) have proved this too. (iii) As soon as we reach a fairly small and manageable remainder (in our mental calculation), we know how many more steps we should expect. Let us again take the case of \ by way of illustration. The first remainder is 3 ; and used as a successive multiplier (with the provision for the casting out of the sevens), that first remainder—multiplier brings us on to 1. When we have done two steps and got 1 and 4 as the first two quotient-digits, we find 2 is the remainder. Multi plying the first group of two digits (14) by 2, we get 28 as the second-group (with the remainder also doubled i.e. 2 x 2 = 4 ). 14/28/. Multiplying 28 by 2, we get 28 x 2 = 5 6 as the third group and 4 x 2 = 8 as the remainder. And then, by casting out the sevens, we obtain 57 as the quotient-group and 1 as the remainder ! And as this was our starting-point, we stop further computations and decide that when decimalised, has 6 decimal places in the answer. Going back to the case of j’y, the student will that, after 4 steps, we got.0588 as the quotient-digits and 4 as the remainder. Multiplying the former by the latter, we obtained 2352 as the second quotient-group and 4 x 4 = 1 6 ; as the remainder ; and there we stopped, (because we had the first 8 digits on hand and knew the other 8 digits). Thus ^T gave us 16 digits.

( 224 ) (As a Geometrical series is of the standard form 1, r, r1 and so on, we are able to utilise 2 and 2a (in the case of \)f 4 and 4* (in the case of fa) and so on for helping us to pre-determinethe number of decimal places in the answer. This is the Algebraical principle utilised herein. N o t e 1.

We need hardly point out that the Ekddhika method has the supreme and superlative merit of lightening our division (and multiplication) work. For ins tance, in the case of fa , fa etc., we have to do our * division-work, at stage after stage, by successive division, not by 19 or 29 etc., (the original deno minator) but by 2 or 3 etc. (the Ekadkika-purva). And this is the case with regard to every case i.e. that we perform all our operation^ —in this system— with much smaller divisors, multipliers etc., and this rule is invariable. What a tremendous saving in effort, labour, time and co s t!

2.

We have purposely treated this subject at great length and in elaborate detail, because it is very essential that the whole matter should be clearly understood, thoroughly assimilated and closely followed so that, even without the help of a teacher, the student may be enabled to work out these methods independently in other similar cases and to know—with absolute certainty—that ANY and EVERY vulgar fraction can be* readily tackled and converted into the corresponding tecurring decimal {whatever may be the complexity thereof and the number of decimal places therein). In fact, in as much as these simple and easy processes are available—and suitable-for ALL possible denominators and for all possible numerators, the decimal (arid especially the recurring decimal) should no longer be a bugbear to the student. On the contrary, they should be the most welcome of all welcome friends !

( 225 ) Some Characteristic features (General and Special) (1) In the cases of fractions with prime numbers (like 19, 29, 59 etc.,) as denominators, the maximum number of decimal places is one less than the denominator ! This is self-evident and requires no elaboration. (2) Usually, it, or a sub-multiple thereof, is the actual number. (3) Generally, the rule of complements (from nine) is found in operation amongst them. (4) For fractions like fa fa , fa etc., (where the denominator are products of prime numbers), the number of digits depends on the various respective factors in each case (as will be presently elucidated). (5) If and when the decimal-fraction obtained from one of the factors of the denominator is exactly divisible by the other factor (or factors), the division by the second factor leaves no remainder. And therefore the number of decimals obtained by the first factor is not added to ! Thus, (iv I *142857 U 7 X 3 ““ 3 Here, the numerator on the R. H. S. being exactly divisible by 3, it divides out and leaves no remainder. There fore, the number of digits continues the same. This means that, in every case wherein the complementary halves (from nine) are found, the numerator on the R. H. S. must necessarily be divisible by 3, 9 etc. And by multiplying the denominator in such a case by such factors, we cause no difference to the number of decimal places in the answer. And consequently, we have : - oi 5873 ; and so on. fa= z—1 7X9 9 Going back to the Ekanyuna Sutra (as explained in con nection with the Sanskrit Alphabetical code), we know that 142857=143X999=11 X 13 X33X 37. This means that since the numerator is divisible by 11,13, 3, 9, 27, 37, 33, 39, 99, 117, 297, 351 and 999, the multiplication of the Denominator (7) 29

(

226

)

by anyone of these factors will make no difference to the number of decimal places in the answer. (ii) V'

13X3

07_ggg3 3

Here too, all the above considerations apply. And, since 76923=77 X 999= 33X 7X 11 X37, therefore these factors (and combinations of factors) will, by multiplying the denominator, make no difference to the number of decimal places. (Note. 999999=999 X 1001 =999 X 7 X 11 X 13).

#

1

- - and comes under the same category (with Zo X w 22 digits in the answer (just like -fe). (7) is a special case and stands by itself. Naturally it should have been expected to provide for 48 places. But, as a matter of fact, it gives only 42; and for a perfectly valid and cogent reason i.e. that, out of the 48 possible multiples, six (i.e. J J, !£•, f §• and f §) go into a different family, as it were^&nd take shape as $, f , § and f ; have their places there as *142857, *285714 and so forth and need no place in the 5 ^ etc., group ! And thus, since 6 go out of the 48, the .remaining 42 for the 42 places actually found in the decimal equivalent of ^ ! This is not a poet’s mere poetic phantasy but a veritable mathematical verity! (8) is, in a way, an exception, as it contains only 13 digits. And, as this is an odd number, the question of the two complementary halves does not arise! 13, however, is a sub multiple of 78 ; and there is no deviation from the normal in this respect. An at-sight-one-line mental method will soon be given for fa (in this very chapter). (9) Similarly has 44 digits and thus conforms to the sub multiple rule. And this implies that, like TV it will need another complete turn of the wheel (in one of its multiples) in order to meet the needs of all the multiples ! (An incredibly easy method will be shown in this very chapter for reeling off the answer in this case).

( 227 ) (10) fa has only two recurririg places (*01); but the whole gamut can be and has been provided for, therewith. (11) In the case of basic fractions ending in 3, the denomina tor is first multiplied by 3 and gives us the Ekadhika, and the last digit in the answer is also 3. (12) fa (like fa) has only two decimal places (*03). (13) fa has only 21 digits. 21 is a sub-multiple of 42 but is odd and gives no scope for the complementary halves. (14) fa has only 13 digits (a sub-multiple but odd) (15) fa has only 41 digits (similarly). (16) fa is special.

Since 7 3 X 1 3 7 = 1 0 0 0 1 and since

10001X9999=99999999

7

™ L= 10001 99999999

==*0136/9863 (by Ekanyuna Sutra) (17) And,

(18) f a

conversely,

x __ 73 ___7 3 x 9 9 9 9 T* T~ 10001 99999999

*00729927

will be discussed a little later.

(19) In the case of fractions whose . denominators end in 7, the last digit is also 7 ; and the Ekadhika is obtained from the denominator multiplied by 7. (20) y and f a have been dealt with in detail already. (21) ^

and-

are

special

(because

2 7 x 3 7 = 9 9 9 ).

their decimal forms are. *037 and *027. (22) f a has 46 digits.

(9o\ i _ 1

1

and has °nly 18 digits.

19 X 3

]

(24) f a has 33 digits (odd) (25) f a has been discussed already (number ^) /o a\

i

1

an(i has 28 digits.

(26) * * “ 29^3 (27) fa has its full quota of 96 digits.

And

( 228 ) (28) In the case of fractions with denominators ending in 1, the Ekddhika comes from the denominator multiplied by 9 : and the last digit is 9. ( 2 » ) tV = -0 9 ( 3 0)

1 and has been discussed under 7. 7x3

(31) / j wiU come up a little later. (32) & is special V 41x271=11111 . *

i _ 271 _271 x 9 *T H il l 99999 ~ '02439 (odd) ___ . 41 _4 lx9 __0 0 3 6 9 lllll 99999 ~99999= 00369

(33) And, conversely, 1 271

/jj\ x _ 1 and has 10 digits. ’ ” 17X3 (35)

has 60 digits.

(36)

has 35 digits (odd)

(37) * = _ L _ = 9 ? l = - 6 l 2 3 « 6 ? 9 v 1

27X3

/oo\ i _ 1 ” ~ i8 x 7

3

<S

Very * * * * * *

number).

has already been discussed under 7 and under 13. And besides, i - 11 „ H X 9 9 9 . . VT 1001 999999 — *010/989 But

But here a big BUT butts in and exclaims; “ Y es; all this is all right in its own way and so far as it goes. But, as our denominators go on increasing, we note that, although the last digit of the decimal fraction is 1,3,7, or at the most 9 and no more, yet, the Ekddhika Purva goes on increasing steadily all the time and we have to multiply or divide successively by bigger and bigger Ekddhikas, until, at last, with only two-digit denominators like 61, 71 and 81 and so on, we have now to deal with 55, 64, 73 etc., as our multipliers and divisors, and surely this is not such an easy process.

( 229 ) The objection is unobjectionable; nay, it is perfectly correct. But we meet it with quite a variety of sound and valid answers which will be found very cogent and reasonable. They are as follows: (i) Even the biggest of our Ekddhikas are nowhere—in res pect of bigness—near the original divisor. In every case, they are smaller. But this is only a theoretical and dialectical answer from the comparative standpoint and does not really meet the intrinsic objection (about the Vedic methods being not only relatively better but also being free from all such flaws altogether)! We therefore go on and give a satisfactory answer from the positive and constructive stand-point. (ii) Even though the Ekadhika is found to be increasingly unmanageably big, yet the remainders give us a simple and easy device for getting over this difficulty. This we shall demonstrate presently. (iii)

The Ekadhika (so far explained and applied) is not

the whole armoury. There are other Auxiliaries too, wherein no such difficulty can crop up. These we shall expound and explain in a subsequent, but sufficiently near chapter of this very volume ; and they will be found capable of solving the problem in toto; and (iv) Above all, there is the CROWNING GEM of all coming up in a near chapter and unfolding before our eyes a formula whereby, however big the denominator may be, we can—by mere mental one-line Vedic arithmetic—read off the quotient and the remainder, digit by digit! This process of “ Straight Division” , we have already referred to and shall explain and demonstrate, in a later chapter, under this very caption “ Straight (or Instantaneous) Division” . In the meantime, just now, we take up and explain the way in which the remainders come to our rescue and solve this particular problem for us.

( 230 ) Let us take first, the cae of *V We know immediately that the last digit of the decimal is 3 and that the Ekadhika is 7. And then we work as follows: (i) Multiplying digit after digit (as usual) by 7, we have : 04347826086 /9 5 6 5 2 1 7 3 9 1 3 6 /3 4 3 1 1 5 2 6 2 or (ii) dividing digit by digit (as usual) by 7, we have : 0 4 3 4 7 8 2 6 0 8 6/9 5 6 5 2 1 7 3 9 1 3 2 3 5 5 14 6 4 6 These are the usual Ekadhika Purva methods. %

But

(iii) we observe in the first chart, after two digits (1 & 3 have been obtained), the next leftward group (39) is exactly three times (the extreme-right-end one) and we can immediately profit by it. Thus 39 gives us 117, out of which we put down 17 and keep 1 to carry over; 17 gives us 51 + 1=52. 52 gives us 156, out of which we set down 56 and keep 1 to carry over. 56 gives us 168+1 = 169. of these, we put 69 down and keep 1 to be carried over ; and so on. In fact, the whole procedure is exactly like the one which we followed from left to right in respect of (=*020408 16 32......). Thus we have: ^ =^= •0434782608, 69, 56, 52, 17, 39, 13 or(iv) if we wish to start from the left end, go on to the right, that too is easy enough.

We note that, the first digits being completed, we get 8 as the Remainder. We can immediately work out this process by multiplying each two-digit group by 8 (as we did in the case of by 2) and frame the following chart : •04 : 32 : 72 : 24 : 08 : 64 : and so on : 2 : 6 : 2 : : 5 : : 34 : 78 : 26 :

: 69 :

These multiplications by 3 to the left and by 8 to the right are easy enough. Aren’t they ?

( 231 ) Let us now take up and try (as promised at an earlier stage). Obviously, the last digit is 7 and the Ekadhika is 33. This is rather unwieldy as a multiplier or as divisor. We should therefore try and see what we can get from the Remainders. We find them to be 10, 6 etc. We can immediately pounce up on this 6 for our purpose and work in this w a y : *02 being the first two digits of the quotient and 6 being our ratio, the next two digits are obviously 12. These X 6 should give us 72 ; but as 4 will be coming over from the right, we add the 4 and put down 76. This should give us 456, of which the first digit has already been taken over to the left. So 56 remains. But this will be increased by 3 (coming from the right) and will become 59—This gives us 57, 44 and 68 for the next three 2digit groups and 08 for the one thereafter. The 08 group of two digits gives us 48 which, with the carried digit becomes 51. This gives us 06 and 36 (which becomes 38). And then we have 28 (turning into 29, then 74 which becomes 78) and so forth. Thus we have : #T=*02 12 76 59 44 68 08 51 382/ 97 8 Here we notice that, exactly after 23 digits, the comple ments (from nine) have begun. So, we can complete the second half and say : 12 76 59 57 44 68 08 51 06 382 97 87 23 40 42 55 31 91 48 93 617 We have thus avoided the complicated divisions by the original divisor 47 and also the divisions and multiplications by the unmanageable Ekadhika 33 ; and, with the easy remainder 6 as our multiplier, we have been able to obtain all the 46 digits of the answer ! This merely shows that these are not cut-and-dried mechanical processes but only rules capable of being applied to the special kind of cases which they are particularly designed to meet and fit into. And, as for a cut-and-dried formula capable of universal application, that too is forthcoming (as already indicated) and will be dealt with, very soon.

( 232 ) Let us now take up fa which, a little earlier, we promised to deal with soon afterwards. In this case, the last digit is 9 ; and the Ekddhika is 28 (which is nearly as big as the original denominator itself)! We should therefore sift the remainders and find a suitable auxiliary therefrom. In this case, we find 7 is the first significant remainder. So, leaving the Ekddhika process oat of for the moment, we may use the Geometrical Progression principle and achieve our purpose thereby (as we did-with 6-in the case of fa). But let us proceed further and see whether a still more easily manageable remainder is available further up. W e ll; we observe :

•032258 (with remainder 2). The actual remainders (in order) are: 10, 7, 8, 18, 25 and 2 ! This suits us most irably, and we proceed further with the help of 2. Thus: •032258/064516/129,032/258064....... But this means that, after only 15 digits (an odd number), the decimal has already begun to recur I So, we simply say : ^ = 032258064516 129 ! What a simple and easy device! Let us now take up The last digit is 7 ; but the Ekddhika will be 68! So, we seek help from the Remainders. They are : 10, 3 etc., and the quotient-digits are ‘0103... So, multiplying each quotient-group (of 2 digits each) by 3 (as we did, by 2, in the case of £v), we g e t: i =-010309 27 81 JL 83 etc. etc ! Let us take one more example (i.e. ? ! ? ) and conclude. The It will surely

last digit is 7 ; but the Ekddhika will be 698 !

not be an enviable task for even the most practised and expe rienced statistician to multiply or divide, at each step, by such a big figure ! We therefore again seek help from the remainders and the Geometrical Progression Rule.

( 233 ) The Quotient “digits are ’ 001 etc., and the successive remainders are 10. 100, 3 etc. ! This means that we should multiply each group of three quotient-digits by 3 and get our answer (to any number of decimal-places). We thus have: ^ = • 0 0 1 : 003 : 009 : 027 : 081 : 243 : 729 : : : : : : 3 732 etc. etc. The Converse operation. Having dealt, in extenso, with the conversion of vulgar fractions into their equivalent recurring decimals, we now take up the CONVERSE process i.e. the conversion of decimals into the equivalent vulgar fractions. We do not, however, propose to go into such a detailed and exhaustive analytical study thereof (as we have done in the other case) but only to point out and explain one particular principle, which will be found very useful in this particular operation and in many subsequent ones. The principle is based on the simple proposition that •9 = 1 = 1 ; • 99= | | = 1 ; -9 9 9 = 1 1 1 = 1 ; and so forth ad infinitum. It therefore follows that all recurring decimals whose digits are all nines are ipso facto equal to unity ; and if a given decimal can be multiplied by a multiplier in such a manner as to produce a product consisting of only nines as its digits, the operation desired becomes automatically complete. For instance, let us first start with the now familiar deci mal *076923. In order to get 9 as the *0 7 6 9 2 3 last digit, we should multiply this by 3. _______1 ^ Setting this product down (*230769), *2 3 0 7 6 9 we find that, in order to get 9 as the 0*7 6 9 2 3 penultimate digit, we should add 3 to ^ 9 9 9 9 —1 the 6 already there. And, in order ------------- ---- to get that 3, we should multiply the given multiplicand by 1. On doing this, we find that the totals (of the two rows) are all nines! So we stop there and argue that, because the given decimal X 13=999999 (i.e. 1), therefore the fraction should be XV In fact, it is like saying 1 3 x = i .\ x ~ so

( 234 ) (2 ) Secondly, let us take the case of -037 and see how this works. Here as the last digit is 7, so, in order to get 9 as the last digit of the product, we should multiply * 0 3 7 it by 7. And, putting 259 down, we should add 4 2 7 to obtain 9 as' the penultimate digit. And, in order 2 5 9 to get that 4 there, we should multiply the multi0 7 4 plicand by 2 . And, on doing so, we find that the product is -999 (— 1 ). Therefore, the fraction 0 '9 9 9 = 1 X 2 7 = l .v x = 21T (3) When we try the case of 142857 ; we find that multi plication by 7 gives us the all-nine product. -999999 ( = 1 ) ; and therefore we say •142857=^ (4) 047619. We first multiply by 1 , see 1 -047 6 1 in the penultimate place, have to add 8 thereto, ________ 2

9 1

multiply by 2 (for getting that 8 ) and thus find 0 . 9 5 2 3 g 9 that the required answer is fa. — =--------------M -9 9 9 9 9 9 = 1 (5 ) Similarly, we may take up various other (including the long big ones like the equivalents of 1 1 T, t V tV purpose achieved.

*V>

decimals

etc-> an(f invariably we find our

(6 ) But, wha't about decimals ending in even numbers or ? Well; no integral multiplier can possibly get us 9 as the last digit in the product. And what we do in such a case is to divide off by the powers of 2 and 5 involved and use this new method with the final quotient thus obtained. Thus, if We have to deal with 285714 we divide it off by 2, get -142857 -2)-285714 as the quotient and find that multiplication thereof •142857 by 7 gives us the product -999999=1. And therefore we say : 5

X7=l

x=*

(7) Let us now try the interesting - 0 1 2 3 4 5 6 7 9 decimal. 012345679. On applying this new 8 \ method, we find that multiplication by 81 "0 1 2 3 4 5 6 7 9 gives us 1 as the product x= '9 8 7 6 5 4 3 2 •9 9 9 9 9 9 9 9 9 =

1

( 235 ) N.B.

1.

The student should also make use of the Ekanyuna formula. This is readily applicable in every case of “Complementary halves” (including tV tV tV etc. Thus. "i4 9 .gff7 —143x999__l l X l3 999 999 1001 77x999 Similarly. 076923 X 999

11x13 11X13X7

T

; and S0 0n

2.

Similarly, with regard to other factors too, it goes without saying that the removal, in general, of common factors (from the decimal and the denomi nator) facilitates and expedites the work.

3.

The subsequent chapters on “ AUXILIARY FRAC TIONS” and “ DIVISIBILITY” etc., will expound and explain certain very simple and easy processes by which this work (of arithmetical factorisation) can be rendered splendidly simple and easy ; and

4.

Above all, the forthcoming “ Straight Division” method will not merely render the whole thing simple and easy but also turn it into a pleasure and a delight (even to the children).

Some Salient Points and Additional Traits. Thus, the Ekddhika process (forwards and backwards) and the Geometrical Progression relationship between the remainders have given us the following three main principles : (i) The quotient-complements (from 9 ); (ii) The remainder-complements (from the Denominator) ; & (iii) The multiplication of the Garamanka (last digit) of the remainders by the Caramdnka (the last digit) of the decimal, for obtaining each digit of the quotient. Now, apropos of and in connection with this fact, the following few important and additional traits should also be observed and will be found interesting and helpful:

( 238 ) (1) In the case of the remainders are (I), 10, 5, 12, 6 3, 11, 15, 17, 18/9, 14, 7, 13, 16, 8, 4, 2, 1 and the quotient digits are: 0 5 2 6 3 1 5 7 8/9 4 7 3 6 8 4 2 1 . (i) Each remainder by itself, if even; and with the addition of the denominator, if odd, is double the next remainder. This follows from the Ekadhika being 2. (ii) Each quotient-digit is the last digit of its corres ponding remainder. This is because 1 is the last digit of the decimal. (2) In the case o f2V the remainders are : (1) 10,13, 14, 24, 8, 22, 17, 25, 18, 6, 2, 0, 26, 28/19, 16, 15', 5, 21, 7, 12, 4, 11, 23, 27, 9, 3, and 1. (i) The quotient-digits are the last digits thereof (for the same reason as above) (ii) Each remainder (by itself or in conjunction with the denominator or double of it)= thrice its successor; and (iii) Each remainder plus its successor’s successor=the next remainder thereafter. Thus 10+14=24 ; 13+ 24—29= 8 ; 1 4 + 8 = 2 2 ; 24+22 - 2 9 = 1 7 ;8 + 1 7 = 2 5 ; 22+25—2 9 = 1 8 ; 1 7 + 1 8 - 2 9 = 6 ; 25 + 6 —2 9 = 2 ; 1 8 + 2 = 2 0 ; 6 + 2 0 = 2 6 ; and so on 1 N.B.

Note the casting off of the denominator all through. (3) In the case of

the quotient-digits a re:

0112 3595 5056 1797 752 808 i 9887 6404 4943 8202 247 191 ) and the remainders are (1), 10, 11, 21, 32, 53, 85, 49, 45, 5, 50, 55, 16, 71, 87, 69, 67, 47, 25, 72, 8, 80, 88, 79, 78, 68, 57, 36, 4, 40, 44, 84, 39, 34, 73, 18, 2, 20, 22, 42, 64, 17, 81, 9 and 1. (Note the Ratio 9 : 1 . ) The remarkaole thing here is that the numerator+the first remainder=the second remainder and that all through, the sum of any two consecutive remainders is the next remainder thereafter! Thus 1 + 1 0 = 1 1 ; 1 0 + 1 1 = 2 1 ; 1 1 + 2 1 = 3 2 ; and so on to the very end.

{ 23T ) The general form herefor is a, d, a + d , a+ 2d , 2a+3d, 3a-f 5d, 5a+8d etc. The student who knows this secret relationship (between each remainder and its successor) can reel the 44 digits of the answer off, at sight, by simple addition ! (4) In the case of

the remainders are :

(1) 10, 21, 52, 46, 65, 18, 22, 62, 67, 38, 64, 8 and 1. The general form herefor is a, d, a+ 2d , 2a+5d etc. Know ledge of this relationship will be of splendid practical utility in this case. (Note the Ratio 8 : 1.) (5) In the case of

the remainders are :

(1) 10, 31, 34, 64, 19, 52, 37, 25, 43, 16, 22, 13, 61, 58, 28, 4, 40, 55, 67, 49, 7 and 1. (Note the Ratio 7 : 1.) The general form herefor is obviously a, d, a-J-3d, 3a+10 d, 10a*4”33d etc. (6) In the case of 59, the remainders a re: (1) 10, 41, 56, 29, 54, 9, 31, 15, 32, 25, 14, 22, 43, 17, 52, 48, 8, 21, 33, 35, 55, 19, 13, 12, 2, 20, 23, 53, 58/49, 18, 3, 30, 5, 50, 28, 44, 27, 34, 45, 37, 10, 42, 7, 11, 51, 38, 26, 24, 4, 40, 46, 47, 57, 39, 36, 6 and 1 (Note the Ratio 6 : 1). Here the general form is a, d, a+4d> 4a+17d etc. Inductive conclusion

Having thus examined the cases of note the following :

^ and

we

(i) In every case, we start with 1 (the basic numerator) as a sort of pre-natal remainder (which is perfectly justified because we are dealing with a recurring decimal); and we call it a ; (ii) In every case, the first actual remainder is 10; and we call it d ; (iii) And then the successive remainders are a, d, a + d , a + 2d, a + 3d , a-f4d respectively (wherein the co efficient of x is obviously the deficit of the penultimate digit from 9 !

( 238 ) Thus for fa , we have a-f-ld ; for f a , we have a + 2d ; for fa, w~e have a + 3d ; f°r

we have a+ 4d ; and so on.

(iv) And this relationship is maintained systematically all through. In other words, each remainder+the next one or double that or thrice that etc.=the further subsequent remainder. Arguing thus, let us try fa As 3 is 6 less than 9, the general form should be a+6d. This means 1, 10, 61 (i.e. 9), 64 (i.e. 12). 3, 30 (i.e. 4) and 27 (i.e. I). And we find this to be actually correct. (v) And, in case the penultimate digit is more than 9, we should react by subtracting d (and not add to it) at the rate of 1 for each surplus. Thus, our chart will nowread-a, d, a-—d, d- (a—d) i.e. 2d—a, and so on. For instance, for the remainders will be (1)10, —9, 19,—28, 47 and so on. (vi) And, over and above all these details which are different for different numbers (as explained above), there is one multiplier (namely 10) which is applicable to all cases ! And thus, whatever fraction we may be dealing with, 2, 4, 5, 8 or any remainder what soever can be safely put in into the next place with a zero added ! The student will observe that, in all the examples dealt with hereinabove (not only in this particular subsection), every such remainder of two digits (ending in zero) has been invariably preceded by the same number (without the zero) ! With the help of this rule applicable in all cases and the special rules (about d, 2d, 3d, 4d etc.,) ened for the different individual cases, the student should easily now be in a position to make a list of the successive remainders in each case and therefrom, by Caramdnha multiplication, put down the succe ssive quotient-digits without further special labour !

( 239 ) These and many more interesting features there are in the Vedic decimal system, which can turn mathematics for the children, from its present excruciatingly painful character to the exhilaratingly pleasant and even funny and delightful character it really bears ! We have, however, already gone into very great details; and this chapter has already become very long. We therefore conclude this chapter here and hold the other things over for a later stage in the student’s progress.

Chapter X X V I I STRAIGHT DIVISION We now go on, at last, to the long-promised Vedic process of STRAIGHT (AT-SIGHT) DIVISION which is a simple and easy application of the tJRD H VA-TIRYAK Sutra which is capable of immediate application to all cases and which we have repeatedly been describing as the “ CROWNING GEM of all” for the very simple reason that over and above the universality of its application, it is the most supreme and superlative mani festation of the Vedic ideal of the at-sight

mental-one-line

method of mathematical computation.

Connecting Link In order to obtain a correct idea o f the background, let us go back, very briefly, for a very short while, to the methods which we employed in the earlier chapters on division ; and let us start with the case of According to the first method

According to the second

(under the Nikhilam etc., Sutra)?

method (by Pardvartya for-

our chart will read as follows :

mula),, we say :

73 27

3

8 9 : 6 21: 28:

8

98 : 116406: •

;

73 : 3 27 :

2: Q = 534i I 1

: and R = 0 j[

3 14 58: (26) 28 :

8

9

2

8

9—9 51 — 51 153-“ 153

33:

: 3 17 51 5

3

4

23

949

0

0

534 :(36) 10 We have felt, and still feel, that even these comparatively short, intellectual and interesting methods are cumbrous and clumsy (from the Idealistic Vedic standpoint).

And hence the

clamant need for a method which is free from all such flaws and

( 241 ) which fulfils the highest Idealistic ideal e f the Vedic Sutras. And that is as follows: Out of the divisor 7 3 , we put down

3.: 38 9 8 : 2 : ? : 3 3 : 1 :

only the first digit (i.e. 7) in theJDivisor: 5 3 4 : 0: column and put the other digit (i.e. 3) “ on ' top of the flag” (by the Dhvajanka Sutra), as shown in the chart alongside. The entire division is to be by as explained below :

7

; and the procedure is

As one digit has been put on top, we allot one place (at the right end of the dividend) to the remainder portion of the answer and mark it off from the digits by a vertical line. (i) We divide 38 by 7 and get 5, as the quotient and 3 as the remainder. We put 5 down as the first quotient digit and just prefix the remainder (3) up before the 9 of the dividend. In other words, our Actual secondstep Gross Dividend is 39. From this, we, however, deduct the product of the indexed 3 and the first quotient-dight 5 (i.e. 3 X 5 = 1 5 ). The remainder (24) is our actual Nett-Dividend. It is then divided by 7 and gives us 3 as the second quotient-digit and 3 as the remainder, to be placed in their respective places (as was done in the first step). From 38 (the gross dividend thus formed), we subtract 3 xth e second quotient-digit 3 i.e. 9, get the remainder 29 as our next actual dividend and divide that by 7. We get 4 as the quotient and 1 as the remainder. This means our next Gross Dividend is 1 2 from which, as before, we deduct 3Xthe third quotient-digit 4 (i.e. 1 2 ) and obtain 0 as the remainder. Thus we say : Q is 534 and R is zero. And this finishes the whole procedure; and all of it is one-line mental Arithmetic (in which all the actual Division is done by the simple-digit Divisor 7 )! The Algebraical Proof hereof is very simple and is based on the very elementary fact that all arithmetical numbers are 31

( 242 ) merely Algebraical expressions wherein x stands for ten.

For

instance, 3x2+ 5 x + l is merely the algebraical general expression of which (with x standing for 10) the arithmetical value is 351. ing this, let us try to understand the steps by ^neans of which 3 8 9 8 2 is sought to be divided by 73. Alge braically put (with x standing for 10), this dividend is 38xs+ 9x2+ 8 x + 2 ; and this divisor is 7x+3. Now, let us proceed with the division in the usual manner. When we try to 7x+ 3 ) 38x3+ 9x2+ 8 x + 2 (5 x 2+ 3 x + 4 35x3+ 1 5 x 2 divide 38x3 by 7x, our first quotient-digit is 3x3— 6x2 5x2 ; and, in the first = 2 4 x 2+ 8 x step of the multipli 21x2+ 9 x cation of the divisor 3x2—x by 5x2, we get the =29x+ 2 product 35x3+ 1 5 x 2 ; 28X+12 and this gives us the x— 10 remainder 3x 3+ 9x2— 15x2, which really = 10—10=0 means 30x2+ 9 x 2—15x2 ~ 2 4 x 2. This plus 8x being our second-step dividend, we multiply the divisor by the second quotient-digit 3x and subtract the product 21x2-f 9x therefrom and thus get 3x2—x as the remainder. But this 3x2 is really equal to 30x which (with —x + 2 ) gives us 2 9x-f’2 as the last-step dividend. Again multiplying the divisor by 4, we get the product 2 8 x + 1 2 ; and subtract this 2 8 x + 12, thereby getting x— 10 as the Remain der. But x being 10, this remainder vanishes! And there you have the whole thing in a nut-shell. It will be noted that the arithmetical example just hereabove dealt with (i.e. —793~ ) is merely the arithmetical form of 38x3+ 9 x 2+ 8 x + 2 and the Arithmetical chart has merely 7x-f-3 shown the above given algebraical opera tion in its arithmetical shape (wherein 3 : 38 9 8 : 2 x = 1 0 ) and that, whenever the algebraical ? 3 3 :1 working has taken a remainder-digit *5 3 4 :0

( 243 ) over to the right with a zero added, the arithmetical chart shows that particular remainder prefixed to the dight already there. Thus, where 3xs has been counted as 30x2 and added to the 9x2 already there and produced 39x2 as the result, this algebraical operation has been graphically pictured as the prefixing of 3 to 9 and making it 39 ! And similarly, in the next step o f the division, the remainder 3 is prefixed to the 8 already there; and we have to deal with 38 ; and similarly, at least, the 1 prefixed to the 2 gives us 12 (which the 3X 4 subtracted therefrom cancels o u t)! In other words, the given expression 38x3+ 9 x 2- f 8 x + 2 (with 10 substituted for x) is actually the same as 35x3+ 3 6 x 2-f 37x+12.

And we sa y :

38x3+ 9 x 2+ 8 x + 2 _ 35x3+ 36x2+ 3 7 x + 1 2 _ .s y2 , ^ , 4 And 7x+3 7 x+ 3 graphically, this algebraical operation 3 : 38 9 8 : 2 is demonstrated arithmetically in the ^ 1 3 3 :1 manner shown in the margin. ‘________ _ L _ : 5 3 4 : 0 The procedure is very simple and needs no further ex position and explanation. A few more illustrative instances (with running comments, as usual) will however, be found useful and helpful and are therefore given below : (1) Divide 529 by 23. 3 :5 The procedure is exactly the same ^ and is simple and easy. : 2 (2) Divide 4096 by 64. 4 : 6 :

40

9 : 6 : 4 : 1 ;

2 : 9 ® 3 : 0

(3) Divide 16384 by 128 8 : 16 3 8 : 4 12 : 4 11 : 6

( 244 ) (4) Divide 7632 by 94 (ii) New Nikhilam method or (ii) Newest Vedic method 9 4: 7 6 : 3 2 : 4 : 7 6 3 : 2 : 06 : 0 : 42 : 9 : 4 : 2 : / . Q=81 : : 0 36: *-----R =18 :_________________ : 8 1 : 18: 76 : 4 88 81 : 18 (5) Divide 601325 by 76. 6 : 60 1 3 2 : 5 Here, in the first division by 7, we can 7 : 11 6 2 : 2 put 8 down as the first quotient-digit; but the remainder then left will be : 7 9 1 2 : 13 too small for the subtraction expected at the next step. So, we take 7 as the quotient-digit and prefix the remainder 11 to the next dividend-digit. [N.B. For purposes of reference and verification, it will be a good plan to underline such a quotient-digit (because the chart offers itself for verification at every step and any reconsideration necessary at any stage need not involve our going back to the beginning and starting the whole thing over again)] (6) Divide 3100 by 25.

1

5: 3

2 :

2

2 4

0 : 0 : :: 2 : o

: 1

1

Note— In algebraic terminology, 3100=3x8-|-x2= 2x3-)-9x2+ 8 x + 2 0 and the above example is the arithmetical way of stating that 2x3+ 9 x ay 8xy2+ 2 0 y 3= (2 x -f5 y ) (x2- f 2xy + 4 y 2) (i.e. 25X124=3100) (7) Similar is the case with regard to the division of 38x3-t-9x2+ 8 x + 2 by (x —1), wherein Q =38xa-(-47x+55 and R =57. (8) Divide 695432 by 57. 7 : 6 9 5 4 3 : 2 : 5 : 1 2 1 0 : 3 : 1 2 2 0 0:

32

(9) Divide 3279421 by 53. 3 : 32 7 9 4 2 : 1 : 5 : 2 4 6 5 : 6 : 6 1 8 7 5 : 46

( 245 ) (10) Divide 7777777 by 38 8 : 7 7 7 7 7 7 : 7 : 3 : 1 1 5 7 8 : 7 :

(11) Divide 500001 by 89. 9 : 50 0 0 0 : 1 8 : 10 V 8 : 15

: 2 0 4 6 7 8 : 13 :

:

(12) Divide 37941 by 47. 7 : 37 9 4 : 1 : 4 : 5 3 : 6 : :

8

0

9

:

7 : 12 :

4 -3

4

:

88

(13) Divide 745623 by 79 9 : 74 5 6 2 : 3 : 7 : 11 6 9 : 9 :

(14) Divide 7453 by 79 (te 3 places of decimals) 9 : 74 5 3 0 0 : 7 : 11 6 6 5 0: :

5 6 1 7

9

4

3

8 : 21 :

(15) Divide 710-014 by 39 (to 3 places of decimals)

9 : 7 1 0 . 0 1 4 : 3 : 4 8 2 2 6 4:

2 0:

: 1

8. 2

0

5

5

:

(17) Divide 7 3 by 53 (to 5

(16) Divide 220 by 52 (to 3 places of decimals) 2 :2 2 0 -0 0 0 5 : 2 2 1 4

places of decimals)

5

3 : 7 3 0 0 0 0 : : 2 5 6 4 4: :0 1 3

7

7

3

6 :

(18) Divide 71 by 83 (to 5 places of decimals) 3 : 7 1 0 0 00 : 8 : 7 6 53 : :

0 .85

5

42 . . . :

(19) Divide 1337 by 79 (i) By the New Nikhilam

79 : 21 :

13 2 15 16

(ii) By

the

newest

Yedic

method

method

3 1 10 1

7

:

5

:

: :

52 : 21 :

:

73 :

9 : 7 :

13

3 6

1

6

: 7 : 12 :

73

:

( 246 ) (20) Divide 1681 by 41. 1 : 16 8 1 : N .B .:—The Algebraical form is : 4 : 0 0 0 : ---------------------- 16x2+ 8 x + l , , : 4 1 0 : a-v_i_i 4x+l = 4 x + l (21) Divide 115491 by 137. 7 : 115 13 : 11 8

4

4

9 6

1

2 0

:

or in Algebraical form : 13x+7 : 115x3-f- 4x2+ 9 x + l ( 8 x 2+ 4 x : 104x8—56x2 l l x 3—52xs+ 9 x = 5 8 x 2+ 9x E =104x8+ 108x2+ 67x+ 21 52x2+28x 6x2—I9x+1 #\ Q = 8 x 2+ 4 x + 3 I = 4 1 x+ 1 •(i.e. 843) & R = 0 J 39x+21 2x—2 0=0 (22) Divide 7458 by 127 (to (23) Divide 3517 by 127 (to 3 3 places of decimal) places of decimals) 7:74 5 800 0 0 7: 35 1 7 0 0 0 12 : 14 14 8 7 8 4 12 : 11 13 16 10 13 5

8 *7 2 4 4

2 7-6

9

2

(24) Divide 7031985 by 823 Here, the Divisor is of 2 3: 70 3 1 9: 8 5 3 digits. All the difference 8 : 6 7 5 : 4 4 3 and which this makes to us is 3 3 R =273 8 5 4 4 that, instead of putting one extra digit on top, we put both the extra digits (23) there ; and j we adopt a slightly different modus operandi (on the U R D H V A -TIR YA K lines) in respect of the subtraction- portion of the work. In this instance, for instance, we divide 70 by 8 and set 8 and 6 down in their proper places as usual. Thus, our second Gross Dividend is now 63. From that, we subtract 16 (the product of the first of the flag-digits i.e. 2 and the first quotient

( 247 ) digit i.e. 8) and get the remainder (63—16=47) as the actual dividend. And, dividing it by 8, we have 5 and 7 as Q and R respectively and put them down at their proper places. So now, our Gross Dividend is 71 ; and we deduct, by the UrdhvaTiryaJc rule, the cross-products of the two flag-digits (23) and the two quotient-digits (85) i.e. 1 0 -f2 4 = 3 4 ; and our remainder is 71—34=37. We then continue to divide again by 8 and subtract etc., in the same manner (by cross-multiplication) as (just now explained) by the Vrdhva-Tiryah method (until the last digit of the dividend) is reached. And that finishes the task. And, in other divisions too, irrespective of the number of digits in the divisor, we follow the same method. And, in every case, our actual divisor is of one digit only (or at the most, a small two-digit one (like 12, 16 and so on) which one can easily divide b y ) ! And all the rest of the digits (of the divisor) are hoisted on the flag-top. And this is the whole secret of the “ Straight Division” formula. Note :—If instead of the decimal places in the Quotient, you want the remainder, you can have it in the usual way. In this case, 23 and 44 (by cross-multiplication) give us 20, which (when taken to the right) means 200; and 3 x 4 (the last flag-digit X the last obtained quotient digit) = .1 2 . Subtracting the total of these two (i.e. 212) from 485, we have R =273. (i.e. R = 4 8 5 —200—12=273). Some more instances (of division by three-digit divisions etc.) are cited below : (1) Divide 1064321 by 743 (to (2) Divide 222220 by 735 (to 3 places of decimals) 4 places of decimals) 43: 10 6 4 3 2 1 0 0 : 3 5 : 22 2 2 2 0 0 : 7 : 3 4 4 5 7 7 6 5 : 7 : 1 33 5 3 3 : :

14 3 2 .4 6 4 3

R = 5 2 1 - 1 7 0 —6=345

:

3 0 2-340

R = 3 2 0 - 60—10=250

( 248 ) (3) Divide 8 8 8 by 672 (to 3 places of decimals) 72 : 8 8 8 0 0 : or 732 : 8 : 8 . 6 : 2 3 3 4 2: : : 1 : 1 .3 2 1 5 (4)

28 : 5 : :

:

:

: R =216:

1

or by mere Vilokanam (Inspection) 3 8 1 8 : 2 7 : or 32 : 6 3 8 1 8 : 2 1 1 6 5 : : 5 : 1 0 4 5 : 4

6

1

2

0

8

6

: R =419 :

(5) Divide 13579 by 975 75 : 13 5 ‘ 7 9: 9 : 4 : 11 : :

: :

8

1

3 :

:

1

2

0

8

: •

7 12

: R =419 :

6

(6 ) Divide 513579 by 9 3 9 39 : 51 3 5 :7 9: 9 : 6 12 : 14 :

:

:

5 4

6

:

:

R = 1179—2 6 0 —15=904 R =1479—540—54=594 (7) Divide 7143 by 1171 (i) By the new Paravartya (ii) By the new Pardvartya method (Vinculum) method: 1171 : 7 : 1 4 3: 1171 :7 : 1 4 3: — 1 —7 —i : 7—49—7 : _ i _7 — 1 : : —]4 —21—7 : : 7 : 6 45 4 : _2 -j-3 — l : ? • 13 + 2 5 —4 : : 7 : —1054 : : 7 :—1054 : : 6 ; 117 : : 6 : 117 : (iii) By the newest Vedic method 71 : 71 : 4 3 : Q= 6 11 : : 5 1 : and R = 5 4 3 —543—426=117 : 6 : 1 0 (9) Divide 46781 by 1483 (to ( 8 ) Divide 4213 by 1234 (to 4 3 places of decimals) places of Decimals) 83 : 46 : 7 8 1 0 34 : 42 : 1 3 0 0 9 14 : : 4 9 11 12 3 12 : : 6 4 7 3 2 ! 3: 1 .5 4 5 : 3 : 4 1 4 . 1 0 ( 1 0 ) Divide 3124 by 1532 (to 3 places of decimals) 32 : 31 : 2 4 6 15 : : 1 6 0 :

2 .:

0

4

0

( 1 1 ) Divide .333333 by 1782 (to 3 places of decimals) 82 : 33 : 3 3 3 3 17 : : 16 19 8 :

1

:

8

7

( 249 ) ( 1 2 ) Divide 46315 by 1054 (to 3 places ot decimals). 54 : 46 : 3 1 5 0 10 : : 6 13 10 8 :

4:

3 9

4

(13) Divide 75313579 by 1213 13 : 75 : 3 1 3 5 79 12 : 3 3 11 11 11 :

2

R =1315—310—12 (14) Divide 135791 by 1245 45 : 13 : 5 7 91 12 : : 1 11 4

6 : 2 0

(15) Divide 13579 by 1616 16 : 135 : 79 16 : : 7 :

R = 5 9 1 -2 5 0 - 6 = 3 3 5

3 :

8

(17) Divide 97531 by 1627 27 : 97 5 : 31 16 : 17 : 21 :

:

8

R = 7 7 9 —128 or 690 -3 9 = 65 1

R=491 —405 or 130—44=86

:

8

R = 1 179—344=835

* 1 : 0 9

(16) Divide 135791 by 1632 32 : 135 7 : 91 16 : 7 :5

8

5

or

27 : 97 5 : 31 16 : 1 :3 :

9 :

6

0

:

and R = 3 3 1 —4 2 0 = —89 i.e. Q=59 and R=1538

R = 2 1 3 1 -5 9 3 =1538 (18) Divide 97531 by 1818 18 : 97 5 : 31 18 . 7 : 16 :

5

3:

.. R = 1 6 3 1 -4 5 4 (or 1200^23)=1177

(19) Divide 13579 by 2145 45 : 135 : 79 21 : : 9 :

6

:

And R = 9 7 9 -2 7 0 or 930—221=709

(2 0 ) Divide 135791 by 2525 25 : 135 7 : 91 25 : 10 : 22 : 5 3 : 32

And R =2291—325 or 1980—14 = 1966

( 250 ) (21) Divide 50 X 11 by 439 (to three places of decimals). 39 : 50 1 1 4 : 1 33 ,, . . ,: 11 415

0 6

: : or

41 : 50 : * 1 1 0 : R = 1 9 8 = l8 2 4 : 1 : "222 : or 3 1 1 _129 ~ And=182 : 11 : 415 :

:

(2 2 ) Divide 15X61 by 349 (to three plaees of Decimals). 49 : 15X 6 3 : 3 8

1

475

0 85

10

or

51~: 15 . 6 1 0 : 3 : 3 4 4 :

277

:

A n d R = 3 6 1 —196 or 200—35=165

4 47277

And R = 3 6 1 —196=165

(23) Divide 47 by 798 (to five places of decimals). 98 : '4 7 0 0 0 0 7 : 4 = 121915 058

9 0

(24) Divide 1 1 1 1 by 839 39 : 11 : 1 1 : 8 : : 3 4 : 1 :3

:

or

02 : 4 7 0 0 8 : 4 7 6 :0

5

8

9

0

0

0

or By mere Vilokanam (Inspection)

And R = 3 1 1 -3 9 = 2 7 2 We now extend the jurisdiction of the Sutra and apply it to Divisors consisting of a large number of digits. The principle involved being the same, the procedure ;s also identically the same as in the foregoing examples. And the division by a single digit (or a small two-digit divisor) continues exactly the same. A few illustrative instances are given hereunder: ( 1 ) Divide 7031 -95 by 8231 (to 5 decimal places). 231 : 70 3 1 9 5 0 0 8 : 6 7.5 4 6 9

. Q K A QO

(2 ) Divide 995 311 by 16123 123 : 99 5 : 3 1 1 16 •' 3 ; 13

___

R=13311 —1503 (or JJ200+190 +2)=11808

T ( 251 ) (3) Divide 975 311 by 16231 231 : 97 5 : 3 1 16 : 1 :3 :

0:

6

1

R = 1 5 0 0 + 5 0 + l or 3311 —1860=1451

(4) Divide 975 311 by 16 333 333 : 97 5 : 3 16 : 17 : 16 :

5

9

1

R = 16311—4647 = 11664 or 12100+410+26=11664

:

(5) Divide 975 311 by 18123 123 : 97 5 : 311 18 : 7 : 16 :

5

3

1

11=15000+200+8 (or 16311—1519)=14792

:

(6 ) Divide 995311 by 20321 321 : 99 5 : 311 2 0 10 • 23 ‘ ‘ T?.=2ftlOO —190—7 f=23311'— 3408)=19903 : 4 8 : (7) Divide 997531 by 30321 321 : 99 30 : :

7 9

3

531 28

R=27300 —40—1 (or 28531 —1272)=27259

2

(8 ) Divide 137 294 by 5749 (to 749 5

: 13 : :

3

2 3-

7

6

places of decimals).

2 9 4 0 0 13 13 14 14

8

8

8

370

1

or 6

351 : 13 7 : 1

23 ’

9

2

5

3

88

4 0 0 0 2 5 3 0

1370

0 8

1

( 252 ) (9) Divide 53 247 by 4999 (to five places of decimals). 999 • 53. 2 4 7 0 0 4 : 1 4 9 1 1 14 12 • 10.

0 0

r:

5

5

6

1

5

3

or 3. 2 4 7 0 0 0 : 0 3 2 0 2 1 0 — — ------------------------ ------• 1 0.6 5 1 5 3 ... 5

___ N.B. Better to divide by 50

( 1 0 ) Divide 13 8462 by 39898 (to 3 places of Decimals). 6 2 0 0 9898 : 13 8 4 4 9 13 14 0 3 : :

3, 4

7

0

0 3 or

0 1 0 2

4

: 13. 8 4 6 2 : : 1 2 3 4 : 3 .4

7

03 9 : N.B. Better to divide by 40

(11) Divide 131 by 19799 (to 5 places of decimals). 9799 : 1 3 1 0 0 0 1 7 11 13 19 1 : '• 0 0201 2

0

“ : :

:

6

1 0

1

6

.........

or 10 0 0

3 1

0

6

1

6

16

6

N.B. Batter to divide by

( 1 2 ) Divide 76432 by 67998 (to 5 places of decimals). 7998 : 7 6 4 3 2 0 6 : 1 3 6 7 9 10 : : : :

I

1

7

4

6 1

0

1

.

4

2

1

or 3 2

2

3

0

2

4

0

7

6

0

0

18

3 ......

2 0

( 263 ) (13) Divide -2537 bv 48329 8329 : 2 5 3 7 0 or 4 : 5 5 10 16 : .0 5 2 4

2331 ; .2 6 3 7 0 0 5 : 2 0 3 11

9

: .0

5

2

4 9 ...

(14) Divide 371628 * 1 1 2 by 12734 (to 5 decimal places). 2734 : 371628- 112 1 : 14 8 7 11 8 12 8 11 11 11 N.B. Better 291839 25868 divide by 12 or 334 : 3 7 13 : :

1 11

2

2 14

6 0

9 1

8

3

112 14 3 9

8

00 5

8

6

2 5 8 6 8

N.B. Here we have divided by 13 (15) Divide 4-1326 by 31046 (to 5 decimal places). 1054:4- 1 3 2 6 0 0 1046 4 - 1 3 2 6 0 0 1 1 1 2 4 or 3 : 111111 1 * 3

3

1

1

2

...

:1 - 3 3

11

2

(16) Divide "20014 by 137608 (to 5 decimal places). 37608 : 2 0 0 1 4 0 or 42412: 2 0 0 1 4 1 : 13689 1 : 1 2 1 0 1 4 6

44 ...

: •

1

4

5

...

0

44

N-B. Better divide by 13 (17) Divide .0034147 by 814256321 (to 6 decimal places). 14256321 : 0 0 3 4 1 4 7 8 : 3 2 9 5 :

0

0

0

0

419

N.B. The Vinculum method is always available but will not make much difference. In fact, it may prove stiffer. (18) Divide -200103761 by 9371836211 (to 3 places). 371836211 •2 0 0 1 0 3 7 6 1 9 2 5 7 7 0

2

1

3

5

...

( 254 ) (19) Divide 74. 5129 by 9314 314 : 74 - 5 1 2 9 0 0 0 9 : 2 1 3 0 9 9 9 9 8

-

0

0

(2 0 ) Divide 71324 by 23145 3145 : 7 : 1 3 2 4 2 : : 1 2 4 4 : 3 :

0

8

0

0 0

or

966

145 : 71 : 3 2 : : 2 20

23

...

1

0

:

3 :

0

8

4

18

6

1

(2 1 ) Divide 137426 by 743-2x1 '242 X 80'04 (to 4 places of decimals). 432 13 7 : 4 2 6 0 0 0 7 6 : 7 11 7 5 4 8 184 42 1 2

9

1

6

14

004 8

:

6

1• —1 -

1

1

9

12

15

15

8

8

8

17

4

0

4

9

8

6

0

0

8

6

0

1

7

11

9 (approximately)

Ch a p t e r

XXVIII

AUXILIARY FRACTIONS In our exposition of vulgar fractions and decimal fractions, we have so far been making use of processes which help to give us the exact results in each case. And. in so doing, we have hitherto (generally) followed the current system whereby multiplications and divisions by powers of ten are mechanically effected by the simple device of putting the decimal point backwards or forwards (as the case may be).

Conventional Method For instance, we manipulate the decimal point thus :

(2)39=3^.

( 3 ) i l = LI-

800 8 5 70 7 130 13 (4) 3741 _*3741 . and (5) 97654 __ *0097654 110000 11 ’ 90000000 9 But after this has been done, the other operations—of actual division etc.,—have had to be carried out in the usual manner.

Auxiliary Fractions There are certain Vedic processes, however, by which, with the aid of what we call SAHAYAKS (AUXILIARY fractions), the burden of the subsequent operations is also consi derably lightened and the work is splendidly facilitated.

First Type The first (and commonest) type thereof is a very simple and easy application of our self-same old friend the Ekadhika Purva. And the whole modus operandi is to replace the Deno minator by its Ekadhika (i.e. to drop the last digit and increase the penultimate one by 1 ) and make a consequential alteration in the division-procedure (as in the case of other Ekadhika operations). N.B. ;— The student will that, in these operations, the remainder at each step (of a division) ‘s not prefixed to a series of zeroes from the right-hand side, but to each quotient-digit.

{ 256 ) Auxiliary Fractions'(A.F.) (First type) (1) for

the Auxiliary fraction is J

^=-052631578/947368421 (2) for jg-, the A.F. is J »V = l m ^ “03448275862068/96551724137931 (3) for f£ AF is 3-7/6 (4) for

AF is

(5 ) for I f , A F is V -, ( 6 ) for AF i® T3 (7) for xiff) 'A.® tV ( 8 ) for x i* , A F = * ^ (9) for XYA> A F = i xT^ ( 1 0 ) for TV»V» AF is * ( 1 1 ) for * JJfr, AF ^ ( 1 2 ) for xs5 v/W> AF is ~0TV (13) tor s^VW’ AF is •p-°T5- '(14) for xsWs-???* A F = ? 5 JW i (15) for AF ===i&SiytJZS. (16) for H |||, A F = in the above* cases, the first eight denominators end in a single nine ; the remaining eight terminate in 2, 3, 4, 3, 4, ,6 , 7 and 4 nines respectively. The question now is : Does it stand to reason that the Ekadhika should be the same in * and in (irrespective of the difference in the number of nines) ? That would be tantamount to declaring that the same (signi ficant) numerator (ordividend) with two different denominators (or divisors) will yield the same quotient! And that would be palpably absurd ! Yes ; the objection is perfectly valid ; and the relevant Sutra has surmounted this difficulty beforehand, by providing for groups of quotient-digits (to which the remainder at each stage of the mental division should be prefixed)! And that Solves the whole problem.

Modus Operandi For instance, let us take the sixteenth example supra (namely,

whose A.F. is

and whose denominator

ends in four nines) : Here, F is |||||; and AF is and we have to make (in lieu of 4 9 9 9 9 ) our working divisor. As we have dropped 4 nines from the original denominator and have 5 as our Ekd dhika in the Denominator of the Auxiliary fraction, we have to divide the numerator of the latter in bundles, so to say, of

5

4

digits each by

5

; and, whatever remainder there is, has to be

prefixed not to any particular quotient-digit but to the bundle just already reached. Thus, we take up 2*1863 to start with and divide it by 5. r5) 2 - 1863

W e get *4372 as the first

[,Q = *4372 — R = 3 quotient-group and 3 as the remainder. We prefix this remain der to that group and say : •4372 and we divide this dividend (namely, 34372) 3 by the same divisor 5 ; and we g e t : 3

.4372 : 6874 : 2

5374 4

i.e. 6874 is the second quotient-group ; and 2 is the second remainder, which therefore we prefix to the second quotient group. And we continue this process with as many groups as we need. Thus we have :

*4372, 3

6874, 2

5374, 9074 and so on 4 4

(to any number, or tens, or hundreds or thousands etc., of Deci mal places) ! The PROOF hereof is very simple : OOOl : 2 1863 000

( 288 ) N.B. :—The prefixed remainders are not parts of the quotient but only prefixes to the quotient-group in question and are therefore to be dropped out of the answer! This is a simple method by which we avoid divisions by long big divisors and have small and easy denominators to deal with. The student will note that division by big denominators (with a continuous series of zeroes on the right-hand side) and division by the Ekadhika (with the prefixing of the remainder at each step) yield the same result! And this is why the Auxi liary Fraction scheme has been incorporated for lightening the burden of long big divisions. A few more examples are given below : ( 1 ) Express J5 in its decimal shape. 29 Here, 29

. •*

a

F _*^ 3

A F = *20689655172413 l 79310344827586 J (2 ) F = — v' 89

* A F = — / . E==-79775280898 etc., etc. 9

(3 ) F — 1 7 • AF — V' 139 ” 14

1 7

(±\ F = — • AF — 9 8 K9 179 * * 18 3 . [ ' 43 129 ”

(6) F —17/43—51/129

*

*1 2 3 3 403

iq\ p_53^

^

5

• AF—' 53 799 ' ‘ 8

3 0 215827.........

5474860335, 1955....... F = 023255813953488... 13

,\

•*. F = 395 348^837

m F = — = — ‘ AF= — 1 73 219 ' ‘ 22 W

2

/ - F = - 246 57534......

F = 06 63 32 91 61 45 18... 5 2 7 4 3 1

. ;—The upper row (*06633291614518..... .) is the answer and the lower one (5 2 7 4 3 1...... ) is a mere scaffol ding and goes out.

( 259 )

(9) '

F = ~ 899

•A

1

*

(10) F = T | S 1799

F==

a A F -^ 18

01 66 85 20 57 6 7 1 5 7

••• * = 2

1111 F = l ^ = . — • AF=S V ; 233 699 " f

00 11 11 72 8 7 ... 2 13 15 6

42 91 84 54 93 56 22 6 5 3 6 3 1 2 3 1 ............ 5

17— 444 13999 (1 3 ) f f —

• A F = l 4l i - ' - F = 031 716 551 182... ” * 14 10 1 2 3

97017

'

A F =

0097017

29999999 ’' 3 F = 0032339/ 0010779/ 6670259 etc., etc. 0 2 2

The student will have noted that the denominators in all the above cases ended in 9 (or 3 which could be so multiplied as to yield an easy multiple ending in 9). But what about those ending in 1 which would have to be multiplied by 9 for this purpose and would, therefore, aw already pointed out (in the chapter on Recurring Decimals) yield a rather unmanageable Ekddhika ? is there any provision for this kind of tractions ? Yes ; there is. And this takes us on to the second type of Auxiliary fractions. Auxiliary Fractions (Second Type) If and when F has a denominator ending in 1 , drop the 1 and DECREASE the numerator by unity. This is the required second type of Auxiliary Fractions. Thus, (1 ) (2) (3) (4) (5)

for for for for for

3/61, 36/61, 28/71, 73/91, 2 /1 2 1 ,

A F = 2 /60 — ' f AF=35/'60=3 '5/6 A F = 2 7/7 0 = 2-7/7 A F = 7 2/9 0 = 7 ;2/9 A F = l/1 2 0 = ‘ l / 1 2

(6 ) (7) (8 ) (9)

for for for for

14/131, AF=13/130=1 3/13 1/301, AF=0/300 = -0 0 /3 1/901, A F = 0 /9 0 0 = ‘ 00/9 172/1301, AF=171/1300=1 71/13

( 260 ) (10) for 27 43/7001, AF=2742/7000=2 •742/7 ( 1 1 ) for 6163/8001, AF=6162/8000=6*162/8 ( 1 2 ) for 1768/9001, A F = 1767/9000=1 *767/9 (13) for 56/16001, AF=55/16000= 055/16 (14) for 50/700001, A F = 49/700000= *00049/7 (15) for 2175/80000001, A F=2174/80000000= *0002174/8 (16) for 1/900000001, AF=0/900000000= *00000000/9 Modus operandi The principles, the prefixing (to the individual quotientdigits or to groups of quotient-digits) etc., and other details are the same as in the Ekadhika Auxiliary fraction. BUT the procedure is different, in a very important (nay, vital) particular. And this is that after the first division (or group-division is over) we prefix the remainder not to each quotient-digit but to its COMPLEMENT from NINE and carry on the division in this way all through. An illustrative instance will clarify this : Let F be — AF= - = — 2 31 30 3 (i) We divide 1.2 by 3 and set 4 down as the first quotient digit and 0 as the first remainder. *4 0

(ii) We then divide not 04 but 05 (the complement of 4 from 9) by 3 and put 1 and 2 as the second quotient digit and the second remainder respectively. There fore we have *4 1 0 2 (iii) We take now, not 2 1 but 28 as our dividend, divide it by 3 and g e t: •4 1 9

0

2 1

(iv) Thus, dividing 1 0 by 3 , we have : 4 19 3 9 0 - 2 1 1 and so on, until finally our chart reads : F (i.e. J f ) = * 4 1 9 3 5 4 8 3 7 0 9 6 etc., etc. !

0

2

1

1

1

2

1

2

2

02

2

2

( 261 ) Always, therefore, to take the complement (from 9) of each quotient-digit (and not the quotient-digit itself) for the purpose of further division, subtraction etc. whole secret

This is the

of the second type of Auxiliary Fraction.

Some more illustrative examples are given hereunder :

(1)1=1. F—

.-.A F = ± 0 2 4 3 9/0 0 1 1 3 0/0

(2 ) F = ~ 71

lSo, this is a definite recurring J decimal.

AF=— 7

F = -9 8 5 9 1 5 4 9 2 9 5 7 7 4 6 4 7 8 8 7 3 2 3... 64613362645534356652126 ✓ qvtf_91 W 17l

.

r t J ’O

j,_ 1 0 _ 3 0 . ^ p _ —27 81 * " ^ _ 131

. Ap—

F = '5 3 2 1 6 3 7 4 2 6 5 3 2 10 6 12 7 4 11 15

2 ' 9 8

1 ‘ 3 0

~

3 7 0/3 Evidently a re5 0 2/5 curring decimal.

(with groups of

2

digits)

F = 1 8 6 8 75 89 15 83....... 4 5 6 1 5 3 /«\ T7_14°° . *Tji_13 *99 (with two-digit groups) ( ' 1401 ' ‘ 14 F

= 99 92 8 6 2 2 41 13 12 3 5 3

(7\ -p _ 243 1 6 0 1

. ^ j,_ 2 -4 2 (with groups of two digits) ‘ ‘ _ 16

F = 15 17 80 13 74 14 1 1 61 79....... 2 12 2 11 2 1 9 12 4 (8)

„ _5 __ 15 . »ji __ 14 (with groups of two digits) 67 201 " 2 F = -07 46 26 86 56 0 0 1 1 1

( 262 ) /g\ p _2743 . »p__2-742 (with 3-digit groups) ' 7001 ’ ’ 7 *391 801 171 261 248 393 086 5 1 1 1 2 0 4

1

P roof:

001

:

2

7

- 742 999 5 5391 .\ F = 4 0 2/5 0

9*2 23 OR F — — 77 1001

9

7

(with three-digit groups) ’ *

1

F=* 402/597 (evidently a recurring decimal) (n ) ' 9

29 . AF— 15001 ‘ * /. F =

933

001

13

(with three-digit groups) 15

3

204 6

453 036 etc., etc. 0

(T9 \ p__ 137 # iip_*000136 (with 6 -digit groups) ' 13000001 ‘ * 13 F=

0 0 0 0 1 0

6

: 538460: 727810 : 713245 etc., etc. :9 :0 :4

Other Astounding Applications Yes ; but what about still other numbers which are neither immediately below nor immediately above a ten-power base or a multiple of ten etc., (as in the above cases) but a bit remoter therefrom ? Well; these too have been grandly catered for, in the shape of a simple application of the Anurupya Sutra, whereby, after the pre-fixing of each’ Remainder to the quotient-digit in question, we have to add to (or subtract from) the dividend at every step, as many times the quotient-digit as the divisor (i.e. the denominator) is below (or above) the NORMAL which, in the case of all these Auxiliary fractions, is counted as ending, not in zero or a number of zeroes but in 9 or a series of nines!

( 263 ) For example, let F be and suppose we have to express this vulgar fraction in its decimal shape (to, say, 1 0 places of decimals). Lest the student should have, in the course or these peregrinations into such very simple and easy methods of work, forgotten the tremendous difference between the current method and the Vedic method and thereby deprived himself of the requisite material for the purpose of comparison and contrast, let us, for a brief while, picture the two methods to ourselves side by side and see what the exact position is. According to the Vedic method, the process wholly mental is as follows : F = | f , *■* A.F.=^V6'"* But 6 8 being one less than 69 (the normal ending in 9) we shall have to add to each dividend, the quotient-digit in question.

Thus

(i) when we divide 1 *5 by 7, we get as our first quotient-digit and our first remainder.

2

and

1*5 7

1

*2 1

(ii) our second dividend will not be 1 2 but 1 2 + 2 = 1 4 ; and by division of that by 7 , our second Q and R are 2 and 0 .

*2 2

(iii) our next dividend is 0 2 + 2 = 0 4 ; and this gives us 0 and 4 as Q and R. (iv) our fourth dividend is 40+0, giving us 5 and 5 as our fourth Q and R. (v) So, our next dividend is 5 5+ 5 (= 60 ) ; and our Q and R are 8 and 4.

1

0

*2

2

1

®*

0

*2 2 0 5 1 0 4 5 *2 1

2 0

0 4

5 8 5 4

We can proceed on these lines to as many places of decimals as we may need. And, in the present case (wherein 16 decimalplaces have been asked for), we toss off digit after digit (mentally) and say : F = * 2 2 0 5 8 8 2 3 5 2 9 4 1 1 7 6 etc., etc. 1 0 4 5 4 0 2 3 1 6 1 0 1 5 3 2

( 204 )

Over against this, let us remind ourselves of the current method for answering this question: 68)15-0 (-2205882352941176 etc., etc 136 140 136 400 340 600 544 560 544 160 136 240 204 360 340

—200 136 ~640 612 280 272 80

68 120 68

520 476 440 408 Alongside of this cumbrous 16-step process, let us Once again put down the whole working by the Vedic method and say : If ( I f ) — - 2 2 0 5 8 8 2 3 5 2 9 4 1 1 7 6 4 etc., etc. 1 0 4 5 4 0 2 3 1 6 1 0 1 5 3 2 4

( 265 ) A few more illustrative examples are given hereunder: ( 1 ) Express 101/138 in its decimal shape ( 2 0 places), (i) Routine Method— 138)101• 0(* 73188405797101449275 966 440 414 260 138 1220 1104 1160 1104 560 552 800 690 1100

966 1340 1232 980 966 140 138 200 138 620 552 680 552 1280 1242 380 276 1040 966 740 690 50 34

( 266 ) (ii) Vedic Method. (with one below the normal 139) F = -7 3 1 8 8 4 0 5 7 9 7 1 0 1 4 4 9 2 7 4 etc. 3 2 12 10 4 0 8 10 128 0 0 2 6 6 122 10 6 5 N ote:—More-than-one-digit quotient if any, should be carried over (as usual) to the left. (2 ) Conventional method: 97)73- 0(* 75257731958762886597 etc., etc. 679 510 485 250 194

740 679

560 485

610 582

750 679

280 194 860 776

710 679

840 776

310 291

640 582

190 97 H io 873 570 485 850 776

580 485 950 873 770 679 91

(ii) Vedic (at-sigM) method Data: F — AF = V S (but with 2 below the normal 99). .% Add twice the Q-digit at each step. Actual Working: \ F = 3. 7 5 2 5 7 7 3 1 9 5 8 7 6 2 8 8 6 5 9 7 etc., etc. 1 5 6 5 1 1 9 3 7 5 4 1 8 6 4 4 8 5 7

( 267 ) (3) Express

as a decimal

(2 0

places).

Current method— 127)17 0 (-13385826771653543306 etc., etc, 127 430 381 490 381 7 o9o 1016 740 635 1050 1016 340 254 450 381

860 762

690 635

980 889 910 889

550 508

210 127 830 762 680 635

420 381 390 381 900 882 18

(ii) Vedic at-sight method— y

A F =Y ^ (but with 2 below the normal 129) Double the Q-digit to be added at every step.

.\ F = 1 3 3 8 5 8 2 6 7 7 1 6 5 3 5 4 3 3 0 6 etc. 4 4 10 5 9 1 8 8 7 0 8 5 3 6 4 3 3 0 9 12

( 268 ) (4) Express |||f in decimal form (21 decimal places) (i) Usual method— 8997)5236-0(•581 /971 /76S/367/233/522/285 etc., etc. 44985 73750 71976 What a TREMENDDOUS mass and mess of multi plications, sub-

17740 8997 87430 80973

21010 17994

64570 62979

30160 26991

15910 8997

31690 26991

69130 62979 61510 53982 75280 71976

46990 44985 20050 17994 20560 17994

33040 26991 60490 53982 65080 62979

25660 17994 76660 71976 46840 44985 1855

(ii) Vedic at-sight method : 5236 5-236 F= AF= 8997 9

(but with 2 below the normal 8999 and also with groups of 3 digits at a time). Add twice the Q-digit at every step. F = -581 : 971 : 768 : 367 : 233 : 522 : 285 etc., etc. 7 :4 : 1 :1 ;4 :1 :1

( 269 ) (5) Express § £ f f | as a decimal (16 places) Conventional method 49997)21863'0(* 4372/8623/7174/2302/etc., etc. 199988 186420 149991 364290 349979 143110 99994 431160 399976 311840 299982 118580 99994

What a horrible mess ?

185860 149991 358690 349979 87110 49997 371130 349979 211510 199998 115120 99994 151260 149991 126900 99994 26906

(ii) Vedic At-sigkt method—• .. -p_21863 . i i r _ 2 1 8 6 3 49997 ** 5 (with 2 below the normal 49999 and with groups of four digits each). Add double the Q-digit at every step) 62 71 4 2 30 F = 4372 : 8 5(12) 3 : 6(11) 73 : (12) 2 (10) 2 3 : 1 :4 :4 N .B .: very carefully that the extra (or surplus i.e. left-hand side) parts of Q-digits have been “carried over” to the left. This excess is due to the additional multiplication and can be got over in the manner just indicated. A method for avoiding this difficulty altogether is also available but will be dealt with at a later stage. (6) Express as a decimal (eight places) Current method— 76)17-0 (-22368421 etc., etc. 152 180 152

N.B.—Note 84 :

21

:: 4 : 1 .

280 228 520 456 640 608

Even this is bad enough.

{ 271 ) (ii) Vedic At-sight method: V

.*• A F = ^ 1- (but with 3 less than the normal 79)

Thrice the Q-digit is to he added at every step. F = * 2 2 3 6 8 4 2 1 etc., etc. 1 2 4 4 0 0 0 0 (7) Express

as a decimal

(1 2

places)

(i) Usual method— 59998)17125•0( •2854/2618/0872 119996 512540 479984 325560 299990 255700 239992 157080 119996 370840 359988 108520 59998 485220 479984 523600 479984 436160 419986 161740 119996 41744

( 272 ) (ii) Vedic At-sight method :• ..

y

_17125 . AF„1 -7 12 5 59998 ** 6

(with 1 less the normal and with 4-digit groups) only one Q-digit is to be added. F = *2854 : 2618 : 0872 etc., etc. 1 :0 :4

These examples should suffice to bring vividly home to the student the extent and magnitude of the difference between the current cumbrous methods and the Vedic at-sight one-line process in question. Yes ; but what about other numbers, in general, which are nowhere near any power or multiple of ten or a “ normal” denominator-divisor ending in 9 or a series of nines ? Have they been provided for, too ? Yes ; they have. There are methods whereby, as explained in an earlier chapter (the one dealing with recurring decimals) we can easily transform any miscellaneous or non-descript denominator in question—by simple multiplication etc.,—to the requisite standard form which will bring them within the jurisdiction of the Auxiliary Fractions hereinabove explained. In fact, the very discovery of these Auxiliaries and of their wonderful utility in the transmogrification of frightful -looking denominators of vulgar fractions into such simple and easy denominator-divisors must suffice to prepare the scienti fically-minded seeker after Knowledge, for the marvellous devices still further on in the offing. We shall advert to this subject again and expound it still further, in the next two subsequent chapters (dealing with DIVISIBILITY and the application of the Ekadhika Purva etc., as positive and negative OSCULATORS in that context).

Chaptek X X IX DIVISIBILITY AND SIMPLE OSCULATORS We now take up the interesting (and intriguing) question as to how one can determine before-hand whether a certain given number (however long it may be) is divisible by a certain given divisor and especially as to the Vedic processes which can help us herein. The current system deals wth this subject but only in an ultra-superficial way and only in relation to what may be teimed the most elementary elements thereof. Into details of these (including divisibility by 2, 5, 10, 3, 0, 9, 18* 11, 22 and so on), we need not now enter (as they are well-known even to the mathematics-pupils at a very early stage of their mathematical study.) We shall take these for granted and start with the intermediate parts and then go on to the advanced portions of the subject. The Osculators As we have to utilise the “^ t s ” (Ve$anas =Osculators) throughout this subject (of divisibility), we shall begin with a simple definition thereof and the method of their application. Owing to the fact that our familiar old friend the Ekddhika is the first of these osculators (i.e. the positive osculator), the task becomes all the simpler and easier. Over and above the huge number of purposes which the Ekddhika has already been shown to fulfil, it has the further merit of helping us to readily determine the divisibility (or otherwise) of a certain given dividend by a certain given divisor. Let us, for instance, start with our similar familiar old friend or experimental-subject (or shall we say, “ Guinea-pigs” the number 7. The student need hardly be reminded that the Ekddhika for 7 is derived from 7 x 7 = 4 9 and is therefore 5. The Ekddhika is a clinching test for divisibility ; and the process by which it serves this purpose is technically called Vestana or “ Osculation” . 35

( 274 ) Suppose we do not know and have to determine whether 2 1 is divisible by 7. We multiply the last digit (i.e. 1 ) by the Ekadhika (or Positive Osculator i.e. 5) and add the product (i.e. 5 ) to the previous digit (i.e. 2 ) and thus get 7. This process is technically called “ Osculation” . And, if the result of the osculation is the divisor itself (or a repetition of a previous result), we say that the given original dividend (2 1 ) is divisible by 7. A trial chart (for 7) will read as follows : 14;

x 5 + 1 = 2 1 ; and l x 5 + 2 = 7 .-.YES. 21 (already dealt w ith); 28 ; 8 X 5 + 2 = 4 2 ; 2 X 5 + 4 = 1 4 (already dealt with) 35 ; 5 X 5 + 3 = 2 8 (already dealt w ith); 42 (already dea^t w ith); 49; 9 x 5 + 4 = 4 9 . (Repetition mmns divisibility). 56 ; 6 x 5 + 5 = 3 5 (already dealt w ith); 63 ; 3 x 5 + 6 = 2 1 (already dealt w ith); 70 ; O x 5 + 7 = 7 .\ YES 77 ; 7 X 5 + 7 = 4 2 (already done); 84 ; 4 x 5 + 8 = 2 8 (already ov er); 91 ; 1 x 5 + 9 = 1 4 (already dealt with) ; 98 ; 8 x 5 + 9 = 4 9 (already done); Now let us try and test, say, 1 1 2 . 112; 2 x 5 + 1 = 1 1 ; 1 1 X 5 + 1 = 5 6 .\ YES. OR 2X 5+ 11= 21 YES. 4

We next try and test for 13 ; and we find the repetitions more prominent there. The Ekadhika is 4. Therefore we go on multiplying leftward by 4. Thus, 13 ; 3 X 4 + 1 = 1 3 26; 6 X 4 + 2 = 2 6 39; 9x4+ 3= 39 The repetition etc., is uniformly 5 2; 2 x 4 + 5 = 1 3 there and in correct sequence 65 ; 5 X 4 + 6 = 2 6 too (i.e. 13, 26, 3 9 )! YES. 78; 8 X 4 + 7 = 3 9 91 ; 1 X 4 + 9 = 1 3 104; 4 X 4 + 1 0 = 2 6

( 278 ) Examples of the Osculation Procedure (Vestana) A few examples will elucidate the process: (1) 7 continually osculated by 5 gives 35, 28, 42, 14, 21 and 7. (2) 5 so osculated by 7 gives 35, 38, 59, 6 8 , 62,- 20, 2 and so on. (3) 9 (by 7) gives 63, 27, 51, 12, 15 etc. (4) (5) (6 ) (7) (8 ) (9) (10) (11) (12) (13) (14)

(by 16) gives 128, 140, 14 etc. 15 (by 14) gives 71, 21, 16 etc. 18 (by 12) gives 97, 93, 45, 84, 54, 53, 41, 16 etc. 36 (by 9) gives 57, 6 8 , 78, 79, 8 8 , 80, 8 etc. 46 (by 3) gives 22, 8 etc., etc. 49 (by 16) gives 148, 142, 46, 100, 10, 1 etc., etc. 237 (by 8 ) gives 79, 79 etc., and is divisible by 79. 719 (by 9) gives 152, 33, 30, 3 etc., etc. 4321 (by 7) gives 439, 106, 52, 19, 64, 34, 31, 10, 1 etc. 7524 (by 8 ) gives 784, 110, 11, 9 etc., etc. 10161 (by 5) gives 1021, 107, 45, 29, 47, 39, 48, 44, 24, 22, 12, 11, 6 etc. (15) 35712 (by 4) gives 3579, 393, 51, 9 etc. (16) 50720 (by 12) gives 5072, 531, 65, 6 6 , 78, 103, 46, etc., etc. 8

N.B. :— We need not carry this process indefinitely on. We can stop as soon as we reach a comparatively small number which gives us the necessary clue as to whether the given number is divisible (or not) by the divisor whose Ekadhika we have used as our osculator ! Hence the importance of the Ekadhika. Rule for Ekadhikas ( 1 ) For 9, 19, 29, 39 etc., (all ending in 9), the Ekadhikas are 1, 2, 3, 4 etc. ( 2 ) For 3,13,23,33 etc., (all ending in 3) multiply them by 3 ; and you get 1 , 4, 7, 1 0 etc., as the Ekadhikas. (3) For 7,17,27, 37 etc., (all ending in 7) multiply them by 7 ; and you obtain 5,12,19. 26, etc., as the Ekadhikas.

( 276 ) (4) For 1, 11, 21, 31, etc., (all ending in 1), multiply them by 9 ; and you get 1, 10,19, 28 etc., as the Ekadhikas.

Osculation by own Ekadhika Note that the osculation of any number by its own Eka dhika will (as in the case of 7 and 13) go on giving that very number or a multiple thereof. Thus, (1) 23 osculated by 7 (its Ekadhika) gives 7 x 3 + 2 = 2 3 ; 46 (osculated by 7) gives 7 x 6 + 4 = 4 6 ; 69 (similarly) gives 7 x 9 + 6 = 6 9 ; 92 (likewise) gives 2 x 7 + 9 ^ 2 3 ; 115 (similarly) gives 7 x 5 + 1 1 = 4 6 And so on. Now, 276 (osculated by 7) (by way of testing for divisi bility by 23) gives 7 x 6 + 2 7 = 6 9 which again gives 69! YES. Thus, all the multiples of 23 fulfil this test i.e. of osculation by its Ekadhika (7). And this is the whole secret of the Vestana sub-Sutra.

Modus Operandi of Osculation Whenever a question of divisibility comes up, we can adopt the following procedure. Suppose, for instance, we wish to know—without actual division—whether 2774 is divisible by 19 (or not). We put down the digits in order as shown below. And we know that the Ekadhika (osculator) is 2. (i) We multiply the last digit (4) by 2, add the product (8) to the previous digit 7 and put the total (15) down under the second right-hand digit. (ii) We multiply that ,15 by 2, add that 30 to the 7 on the upper row, cast out the nineteens 2 7 7 4 (from that 37) and put down the 3715 remainder 18 underneath that 7. (18)

N.B. — This casting out of the nineteens may be more easily and speedily achieved by first osculating the 15 itself, getting 11, adding it to the 7 ^ ^ ± (to the left-hand) on the top-row and ig J5 putting the 18 down thereunder.

( 277 ) (iii) We then osculate that 18 with the 2 to the left on the upper row and get 38 ; or we may osculate the 18 itself, obtain 17, add the 2 and get 19 as the final' osculated result. ^ ^ ^ ^ And, as 19 is divisible by 19, we say the given number (2774) is also divisible thereby. This is the whole process ; and our chart says : By 19 ? •\ The osculator is 2

f 2 \

7 18

7 15

4

1 ,\ YES. J

OR Secondly, we may arrive at the same result-as effectively but less spectacularly by means of a continuous series of oscula tions of the given number (2774) by the osculator (2) as here inbefore explained. And we can say : Y 2774 (osculated by the osculator 2) gives us 285. 38 and 19 2774 is divisible by 19. N.B. :—The latter method is the shorter but more mechanical and cumbrous of the two ; and the former procedure looks neater and more pictorially graphic, nay, spectacular. And one can follow one’s own choice as to which pocedure should be preferred. N ote:—Whenever, at any stage, a bigger number than the divisor comes up, the same osculation-operation can always be performed. Some more specimen examples are given below : (1) By 29 ?

The osculator is 3. j 3 I 27

2 8 9 6 ? .*. No. 8 31 27 \

OR E (osculated by 3) gives 3307 351, 38, 27, etc. (2) By 29 ?

The osculator is 3. f 9

3

No.

1 4 81

t 29 26 27 28

YES

J

OR The osculation-results are 9338, 957, 116 and 29 .\ YES

( 278 ) The osculator is 3. I 2 4 3 4 5 2 1 ) YES | 29 9 21 6 20 5- \ OR The osculation-results are 243455, 24360, 2436 261 and 29 YES

(3) By 29 ?

(4) By 39?

The osculator is 4. I 4 9 1 4 \ j 39 38 17 J OR The osculation-results are 507, 78 and 39

YES YES

(5) By 49? / . The osculator is 5. ( 5 3 3 2 ] 51^ 19 13 f .. No. C (10) ) OR The osculation-results are 543, 69, 51 and 1 0 No ( 6 ) By 59? / . The osculator is 6 . f 1 9 1 5 7 3 1 .*. YES I 59 49 46 37 25 J OR The osculation-results are 19175, 1947, 236 and 59 YES (7) By 59 ?

The osculator is

.f 1 2 5 6 7 \.\ YES I 59 49 57 48 J OR The osculation-results are 1298, 177 and 59 YES. 6

(8 ) By 59 ? .\ The osculator is 6 . f 4 0 1 7 9 t 47 17 52 38 15

1

\ J

No.

OR Theosculation-results are 40185, 4048, 452 and 57 /.N o. (9 ) By

7 9

?

The osculator i s 8 . f 6 3 0 9 4 8 2 ll .\ No. 113 70 38 64 76 9 10 J

OR The osculation-results are 6309490, 630949, 63166, 6364, 6 6 8 , 130 and 13 No. ( 1 0 ) By 43 ? .•. The osculator is 13. f 1 4 0 6 ll 1129 119 118 19 J

YES

OR The osculation-results are 1419, 258 and 129 .-. YES ( 1 1 ) By 53? .'.The osculator is 16. f 2 , 1 9 5 3 ) / . No. t 149 39 62 53 J OR The osculation-results ar 2243, 272 and 59 ( 1 2 ) By 179?

No.

The osculator is 18.(7 1 45 5 0 1 ).\ YES 1179109 6 2015018 j

OR The osculation-results are 714568, 71600, 7160, 716 and 179 YES

( 279 ) (13) Determine whether 5293240096 is divisible by 139 (or not) (A) By the current method (just by way of contrast): 139)5293240096(380864 417 1123 1112 1124 1112

1200

1112 889 834 556 556 ~ 0

YES.

(B) By the Vedic method: By 139 ?

The Ekadhika (osculator) is 14.

J5 2 9 3 2 4 0 0 9 6 ? YES 1 139 89 36 131 29 131 19 51 93 \ OB The osculation-results are 529324093, 52932451 5293259 529451, 52959, 5421, 556 and 139 YES Note :—In all the above cases, the divisor either actually ended in 9 or could— by suitable multiplication— be made to yield a product ending in 9 (for the determination of the required Ekadhika or Osculator in each case). But what about the numbers ending in 3, 7 and 1 (whose Ekadhika may generally be expected, to be a bigger number) ? Is there a suitable provision for such numbers being dealt with (without involving bigger Ekadhi]tfi-m\i\xi]A'i
( 280 ) (as in the case o f o f the Ekddhika) but of Subtraction (leftward). And this actually means a consequent alternation, of plus and minus. Examples of the Negative Osculation Process ( 1 ) 36 thus osculated by 9, gives 3—5 4 = —51. (2) 7(osculated by 5) gives 0—3 5 = — 35 (3) 35712 (osculated by 4) will yield 8—3571 = —3563. How toDetermine the Negative Osculator Just as the Ekddhika (the positive Vestand) has been duly defined and can be correctly ascertained, similarly the Negative Osculator will also require to be determined by means of a proper definition and has been so defined with a view to proper recognition. It consists of two clauses : (i) In the case of all divisors ending in 1 , simply drop the on e; and (ii) in the other cases, multiply so as to get 1 as the last digit of the product (i.e. 3 by 7, 7 by 3 and 9 by 9 ); and then apply the previous sub-clause (i.e. drop the 1 ). N ote:—For facility of symbolisation, the positive and the negative osculators will be represented by P and Q respectively. Examples of Negative Osculators ( 1 ) For 1 1 , 2 1 , 31, 41, 51 and other numbers ending in 1 , Q is 1 , 2 , 3, 4, 5 and so on. [Note that, by this second type of oscultors, we avoid the big Ekadhikas (produced by multiplying these numbers by 9)]. (2 ) For 7, 17, 27, 37, 47, 57 etc., we have to multiply them by 3 (in order to get products ending in 1 ). And they will be 2 , 5, 8 , 1 1 , 14, 17 and so on. (In these cases too, this process is generally cal culated to yield smaller multipliers than the multiplication by 7 is likely to do).

t 281 ) ( 3 ) For 3 , 13, 23, 33, 43, 53 etc., we have to multiply them by 7 ; and the resultant Negative Osculators will be 2, 9, 16, 23, 30, 37 etc., (which will generally be found to be bigger numbers than the Ekddhikas). (4) For 9, 19, 29, 39, 49, 59 etc., we have to multiply these by 9 ; and the resultant Negative Osculators will be 8 , 17, 26, 35, 44, 53 etc., (all of which will be much bigger than the corresponding Ekddhikas).

Important and Interesting Feature Note :—A very beautiful, interesting and important feature about the relationship between F and Q, is that, whatever the Divisor (D) may be, P +Q =-D . i.e. the two osculators together invariably add up to the Divisor. And this means that, if one of them is known, the other is automatically known (being the comple ment thereof from the divisor i.e. the Denominator).

Specimen Schedule of Osculators P and Q Number 1

3 7 9 11

13 17 19 21

23 27 29 31 33 37 39 41 43 47 49 51

Multiple for P 9 9 49 W 99 39 119 (19) 189 69 189 (29) 279 99 259 (39) 369 129 329 (49) 459

Multiple for Q (1 ) 21 21

81 (1 1 ) 91 51 171 (2 1 ) 161 81 261 (31) 231 111

351 (41) 301 141 441 (51)

P

Q

1

0

I 5 1 10

4 12 2

19 7 19 3 28

2 2 8 1

9 5 17

Total 1

3 7 9 11

13 17 19

2

21

16

23 27 29 31 33 37 39 41 43 47 49 51

8

10

26 3 23

26 4 37 13 33 5 46

35 4 30 14 44 5

11

( 282 ) Number

Multiple for P Multiple for Q 159 399 (59) 549 189 469 (69) 639 219 539 (79) 729

53 57 59 61 63 67 69 71 73 77 79 81 N.B.

371 171 531 (61) 441

201 621 (71) 511 231 711 (81)

P

Q

Total

16 40

37 17 53

53 57 59 61 63 67 69 71 73 77 79 81

6

55 19 47 7 64

22 54 8 73

6 44

20 62 7 51 23 71 8

It will be noted :— (i) that P +Q always equals D ; (ii) multiples of 2 and 5 are inissible for the purposes of this schedule; viii) and these will have to be dealt with by dividing off all the powers of 2 and 5 (which are factors of the Divisor concerned). A few sample examples (1 ) (2 ) (3 ) (4) (5 ) (6 ) (7 ) (8 ) (9 ) (1 0 ) (1 1 ) (1 2 ) (13) (14) (15) (16)

for for for for for for for for for for for for for for for for

59, 47, 53, 71, 89, 83, 91, 93, 97, 99,

P is 6 Q = 53 Q is 14 P = 33 P is 16 Q = 37 Q s 7 P = 64 P is 9 Q = 80 P is 25 Q = 58 P is 82 Q= 9 Q is 65 P = 28 P is 6 8 Q = 29 Q is 89 P= 10 1 0 1 , P i s 91 Q= 10 103, Q is 72 P = 31 107, P is 75 .\ Q = 32 131, Q is 13 P =118 151, P is 136 Q = 15 2 0 1 , Q is 2 0 P = 181

P + Q --D throughout

( 288 ) N ote:—( 1 ) If the last digit of a divisor be 3, its P < its Q ; (2 ) If the last digit be 7, its Q < its P ; and (3) in the actual working out of the subtractions of the osculated multiples (for the negative osculators), the actual result will be an alternationof plus and minus. Explanation ( 1 ) In the removal of brackets, a series of subtrac tions actually materialises in an alternation of + and—-For example, a _ (b -{c -[d -e = f))] = a —b + c —d + e. Exactly similar is the case here. (2 ) When we divide aw-fb * by (a + b ), the quo tient consists of a series of which are alternately plus and minus. Exactly the same is the case here. Note :— The student will have to carefully this alter nation of positives and negatives. But the better thing will be, not to rely on one’s memory at each step but to mark the digits beforehand, alternately, say, by means of a Vinculum (from right to left), on all the even-place digits, so that there may be an automatic safeguard against the possible playing of any pranks by one’s memory. Armed with this safeguard, let us now tackle a few illus trative instances and see how the plan works out in actual practice. ( 1 ) By 41 ? The (Negative) Osculator is 4 ( 1 6 5 7 6 3 ) i —41 — 1 0 4 31 6 /.Y E S Or The osculation-results are 16564, 1640,164 and 0 .*. YES (2 ) By 31 ?

\ Q—-3

6

J 0 33

“o 3 \ 9 J .- .Y E S

Or The osculation-results are 651, 62 and

0

.\ YES.

( 284 ) (3) By 41?

Q =4

J 1 T 2 3_ 4 \ ; 0 10 13 13 ( .-.YES

Or The osculation-results are 1107, 82 and 0 (4) By 47 ?

.'. Q =14 i 1 4 \ 11 102

2 1 6 7 51 64

YES

5) j .'. NO

Or The osculation-results are 74146, 7330, 733 and 31 .'. NO (5) By 51 ?

Q =5

f 4" 3 T 3 2* 1 ( I - 5 10 32 18 3 J .'. NO

Or The osculation-results are 43727, 4337, 398 & 1 .\ NO (6) By 61? Or

Q =6 f l
1\ J .'. YES

The osculation-results are 1952, 183 and 0 .-.YES (7) By 67?

Q =20 J f 0 f7 f 2 0" 31 J 0 - 1 0 100 5 - 8 1 - 4 60 J .- .Y E S

Or The osculation-results are 1017060, 101706, 10050, 1005 and 0 YES. (8) By 91? Q=9 ( 9 8 0 4 5 9 ' 0 F 3 ? Or I 84 69 49 56 37 44 16 22 \ .-.NO The osculation-results

are 98045878, 9804515, 980406 97986, 97441 938 and 21 ,\ NO

(9) By 61 ? .-. Q = 6 H 2 j j 1 3~ 0 5 ' 4 7 Or ( 0 - 1 0 - 2 0 10 53 19 \ ..Y E S The osculation-results are 1221281, 122122, 12200, 1220, 122 and 0 .\ YES (10) By 71 ? Or

Q=7

(“8 0 9 1 0 62 19

0 4 4 31

5 ? \

YES

The osculation-results are 80869, 8023, 781 and 71 .\ YES (11) By 131 ? .*. Q =13 J 1 3 0r 1 0 10

3 J 9 (T 3 ? 1 20 123 39 J

YES

The osculation-results are 133751, 13362, 1310 & 131.'.YES (12) By 141? .'. Q =14 f 4 8 9 ? 8 ~5 7 ? \94 87 37 2 41 93 J .'. NO N.B. But this dividend (yielding the same results) is divisible by 47 (whose Q is also 14). (9 4 = 4 7 x2 )

Chapter X X X DIVISIBILITY ANB

COMPLEX MULTIPLEX OSCULATORS The cases so far dealt with are of a simple type, involving only small divisors and consequently small osculators. What then about those wherein bigger numbers being the divisors, the osculators are bound to be correspondingly larger ? The student-inquirer’s requirements in this direction form the subject-matter of this chapter. It meets the reeds in question by formulating a scheme of groups of digits which can be osculated, not as individual digits but in a lump, so to say. Examples of Multiplex Vestana i.e. (Osculation) ( 1 ) 371 oscillated by 4 for 2 digits at a time, gives 3 + 71 X 4 (=287) and 3—284 ( = —281) for plus oscillation and minus oscillation respectively. (2) 1572 osculated by 8 for 2 digits gives 15+576 (=591) and 15—576 ( = —561) respectively. (3) 8132 osculated by 8 (P and Q) for 2 digits gives 81 +256 (=337) and 81—256 ( = —175) respectively. (4) 75621 osculated by 5 (P and Q) for 3 digits gives 75+3105 (=3180) and 75—3105 ( = —3030) respec tively. (5) 61845 osculated by 7 (P and Q) for 3-digit groups gives 61+5915 (=5976) and 61—5915 ( = —5854) respectively. (6) 615740 osculated by 8 (P and Q) for 3-digit packets gives 615+592C (=6535) and 615—5920 ( = —5305) respectively. (7) 518 osculated by 8 (P and Q) for 4-digit bundles gives 0+4144 (=4144) and 0—4144 ( = —4144) respectively. (8) 73 osculated by 8 (P and Q) for five-digit groups yields 0+584 (=584) and 0—584 ( = —584) respectively.

( 286 ) (9) 210074 osculated by 8 (Pand Q) for five-digit bundles give 2+80592 (=80594) and 2—80592 ( = —80590) respectively. ( 1 0 ) 7531 osculated by 2 (P) for 3 digits gives 7+1062=1069 ( 1 1 ) 90145 osculated by 5 (Q) for 3 gives —7 2 5 + 9 0 = —635 (12) 5014112 osculated by 7 (Q) for 4 gives 501—28784 (13) 7008942 osculated by 3 (P) for (14) 7348515 osculated by

8

(P) for

2

= —28283 gives 126+70089

3

=70215 gives 7348+4120 =11468

(15) 59076242 osculated by 7 (Q) for

2

gives —590762+294 = —590468

Categories of Divisors and their Osculators. In this context, it should be noted that, as there are various types of divisors* there are consequent differences as to the nature and type of osculators (positive and/or negative) which will suit them. They are generally of two categories: (i) those which end in nine (or a series of nines) [in whch case they come within the jurisdiction of the • Ekadhika (i.e. the Positive) Osculator] or, which terminate in or contain series of zeroes ending in 1 , (in which case they come within the scope of operations performable with the aid of the Viparita (i.e. the negative) osculator; and (ii) those which, by suitable multiplication, yield a multiple of either of the two sorts described in sub section (i) supra and can thus be tackled on that basis. The First Type. We shall deal, first, with the first type of divisors, namely, those ending in 9 (or a series of nines) or 1 (or a series of zeroes ending in unity) and explain a technical terminology and symbology which will facilitate our operations in this context.

( 287 ) ( 1 ) Let the divisor be 499. It is obvious that its osculator P is 5 and covers 2 digits. This fact can be easily expressed in symbolical language by saying : P 2= 5 (2 ) In the case of 1399, it is obvious that our osculator (positive) is 14 and covers 2 digits P 2=14. (3 ) As for 1501, Q obviously comes into play, is 15 and covers 2 digits. In other words, Q2=15. (4 ) For 2999, P is 3 and covers 3 digits .*. P 3 = (5) (6 ) (7 ) (8 ) (9) (1 0 ) (1 1 ) (1 2 )

For For For For For For For For

5001, Q8= 5 7001, Q8= 7 79999, P 4 = 8 119999, P 4 = 1 2 800001, Q6 = 8 900001, Q6= 9 799999, P 6 = 8 1 2 0 0 0 0 0 0 1 , Q7 =

3

1 2

Correctness of the Symbology The osculation-process invariably gives us the original number itself (or a multiple thereof) or zero : For example, (i)

4 9 9

(with P a= 5 ) gives us 4 + 5 (99)=4+495=499

(ii) 1399 (with P 2= 14) gives 13+14 (99)=13+1386=1399 (iii) 1501 (with Q2=15) gives 1 5 x 1 —15=0 (iv) 2999 with P 8 = 3 ) gives 2 + 3 (999)=2+2997=2999 (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)

5001 (with Qs= 5 ) gives 5 X 1 —5 = 0 7001 (with Q8 = 7 ) gives 7 x 1 —7 = 0 79999 (with P 4 = 8 ) gives 7 + 8 (9999)=79999 119999 (with P 4 = 1 2 ) gives 1 1 + 1 2 (9999)=119999 800001 (with Q 5 = 8 ) gives 8 x 1 —8 = 0 900001 (with Q5 = 9 ) gives 9 x 1 —9 = 0 799999 (with P 6 = 8 ) gives 7 + 8 (99999)=799999 1 2 0 0 0 0 0 0 1 (with Q7 = 1 2 ) gives 1 2 x 1 — 1 2 = 0

N .B. :—The osculation-rule is strictly adhered to ; and the P’s and the Q’s invariably yield the original dividend itself and zero respectively!

( 288 ) Utility and Significance of the Symbology The symbology has its deep significance and high practical utility in our determining of the divisibility (or otherwise) of a certain given number (however big) by a certain given divisor (however large), inasmuch as it throws light on ( 1 ) the number of digits to be taken in each group and (2 ) the actual osculator itself in each individual case before us. A few simple examples of each sort will clarify th is: ( 1 ) Suppose the question is, Is 106656874269 divisible by

4 9 9

%

Here, at sight, P 2 = 5 . This means that we have to split the given expression into 2 -digit groups and osculate by 5. Thus, f 10 \ 499

6 6

497

56 186

87 525

42 387

69 7 \ /.Y E S

(69 X 5=345 ; 345+42=387 ; 435 + 3 + 8 7= 5 2 5 ; 5 x 2 5 + 5 + 5 6 = 1 8 6 ; 5 x 8 6 + 6 6 = 4 9 7 ; 5 x 9 7 + 4 + 1 0 = 4 9 9 !) Or The osculation-results are 1066569087,

10666125,

106786, 1497 and 499 .*. YES (2 ) Is 126143622932 divisible by 401 ? Here Q2= 4 \ 12 I — 16

61 400

43 62 185 458

29 99

32 1 J

NO

Or The osculation-results are 125915, 1199 and —385. NO

1261436101,

12614357,

( 3 ) Is 69492392 divisible by 199 ? Here P 2 = .*. ( 69 I 199

2

49 463

23 207

92 \ \

YES

Or The osculation-results give 695107, 6965 and 199 YES

( 289 ) (4) Is 1928264569 divisible by 5999 ? Here Pa= 6 1 5999

928 4999

264 3678

569

Or The osculation-results are 1931678 and 599

YES

(5) Is 2188 6068 313597 divisible by 7001 ? Here Q a =

7

< 21 | 0

8 8 6

-3

068 6127

313“ 3866

597 ) \

YES

Or The osculation-results give 21003 and 0 .

21

064 134, 21885126,

8 8 6

YES.

( 6 ) Is 30102 1300602 divisible by 99 ? Here Pa= l As P2= l and continuous multiplications by 1 can make no difference to the multiplicand, the sum of the groups will suffice for our purpose : v 30+10+21+30+06+2=99

.-. YES.

The second method amounts to the same thing and need not be put down. (7 ) Is 2130

1 1 0 2

1143 4112 divisible by 999 ?

Here P8= l and v 2+13 0 + 1 10 + 2 11 + 4 3 4+ 1 1 2= 9 9 9 (by both methods) YES. (8 ) Is 7631 3787 858 divisible by 9999 ? Here P4= l and v =763+1378+7858=9999 .\ (By both methods), YES. (9 ) Is 2037760003210041 divisible by 9999 ? Here P4= l and v 2037+7600+0321+0041=9999 (By both methods) YES. 37

( 290 ) ( 1 0 ) Is 5246 7664 0016 201452 divisible by 1 0 0 1 ? Here Qs= l ; and v S==524+676+648+016—201+452=221=13 X 17 Divisible by 13 But 1001=7X11X13 .*. Divisible by 13 but not by 7 or by 1 1 . /. NO The Second Category The second type is one wherein the given number is of neither of the standard types (which P and Q readily and instan taneously apply to) but requires a multiplication for the transformation of the given number to either (or both) of the standard forms and for the ascertaining of the P and Q (or both) suitable for our purpose in the particular case before us. The Process of Transformation In an earlier chapter (the one on “ Recurring Decimals” ) we have shown how to convert a given decimal fraction into its vulgar-fraction shape, by so multiplying it as to bring a series of nines in the product. For example, in the case of *142857, we had multiplied it by 7 and got *999999 (= 1 ) as the product and thereupon argued that, because 7 x the given decimal = 1 , that decimal should be the vulgar fraction \ (1 )

*142857X7 = *999999=1 A

Similarly, with regard to *076923, we had multiplied it by 3 (in order to get 9 as the last digit of the product); argued that, in order to get 9 as the penultimate digit, ^ ) •076923 we should add 3 to the already existing 6 13 there and that this 3 could be had only by : multiplying the original given decimal by I 1; 0*76923^ then found that the product was now a series of nines; and then we had argued that, 999999=1 V 1 3 x = l, x must be equal to fa. And x==rs we had also given several more illustrations of the same kind (for demonstrating the same principle and process).

( 291 ) As P (in the present context) requires, for osculation, numbers ending in 9 (or a series of nines), we have to adopt a similar procedure for the same purpose; and, in the case of Q too, we have to apply a similar method for producing a number which will terminate in 1 or a series of zeroes ending in 1. The Modus Operandi A few examples of both the kinds will elucidate the process and help the student to pick up his P and Q. And once this is done, the rest will automatically follow (as explained above). ( 1 ) Suppose the divisor is 857. v 857x7=5999, we can therefore at once say : P3= 6. The test and proof of the correctness hereof is that any multiple of the divisor in question must necessarily fulfil this condition i.e. on osculation by P3, must yield 857 (or a multiple thereof). For instance, let us take 857x13 = 11141. As P 3 = 6 1 1 + 6 (141)=857 ! And this proves that our osculator is the correct one. (2 ) Let us now take 43.

v 43x7= 301,

Q2 = 3 .

Taking 4 3 x 3 (=129) for the test, we see 129 yields 2 9 x 3 —1 = 8 6 ; and 8 6 is a multiple of 43 (being exactly double of it). So, our Q is correct. The significance of this fact consists in the natural consequence thereof, namely, that any number (which is really divisible by the divisor in question must obey this rule of divi sibility by the P process or the Q process. N.B. :— what has already been explained as regards P or Q being greater. In this very case (of 43), instead of multiplying it by 7, getting 301 as the product and ascertaining that Q2= 3 is the osculator, we could also have multiplied the 43 by 3, got 129 as the product, found Px=13 to be the positive osculator and verified it. Thus, in the case of 4 3 x 2 = 8 6 , 8+6(13)=86 P1=13 is the correct positive osculator.

( 292 ) Multiplication by 13 at every step being necessarily more cumbrous than by 3, we should naturally prefer Qa= 3 (to P -^13). In fact, it rests with the student to choose between P and Q and (in view of the bigness or otherwise of multiplier-osculator etc.) decide which to prefer. (3) Ascertain the P and the Q for 137. 137 P s= 3 7 137 27 103 959 274

411 1370

3699

14111

Qs= 14

Obviously Qs= 1 4 is preferable (to P#=37). (Test:

137X8=1296

V Q, gives 1296-0=1296 !)

(4) Determine the P and the Q for 157. v 157X7=1099 /. P a= l l And 157x93=14601 Qa=146. P 2=t=ll is to be preferred (Test:— 157x7=1099 / . P a gives 10+1089=1099) (5 ) Find out the P and Q for 229 V 229X131=29999 P4= 3 This Osculator being so simple, the Q need not be tried at all. But on principle, V 229 x 69=15801 at all. But on principle, 229x69=15801 Q j=159 (obviously a big multiplier) (Test for P4= 3 229x100=2/2900

P=

3

gives 8702=229x38)

( 6 ) Find P and Q for 283 Y 283X53=14999 P „=15 and Y 283X47=13301 Qs=133 P8= 1 5 is preferable (Test 2 8 3X 4 = 1 1 3 2 ; 1+ 1 5 (132)=1981 = 2 8 3 X 7 )

( 293 ) (7) Find P and Q for 359 V 359X61=21899 Pa=219 and V 359X339=14001 Q8= 14 Obviously Qs= 1 4 is to be preferred. (Test:

(i) 359x3=1077

Q j= 1 4 gives 14x77—1 =1077

and (ii) 359X115=41285

Q ,= 14 give 1 4 x 2 8 5 -4 1 =3949=359X11)

(8 ) Ascertain P and Q for 421 V 421X19=7999 Ps = 8 and ••• 421X81=34101 Q2=341 obviously P 8 = 8 is the better one (Test:

421X5=2105

P8 =

8

gives 2+840=842 = 421X 2)

(9 ) Determine P and Q for 409 v 409 X 511=208999 .\ Ps=209 and v 409x489=200001 Q5 = 2 Obviously the Q osculator is preferable. (T est: 409 X 1000=4/09000 Q s=2 gives 18000—4=17996=409 x 44) Having thus studied the multiplex osculator technique and modus operandi, we now go on to and take up actual examples of divisibility (which can be easily tackled by the multiplex osculatory procedure). Model Applications to Concrete Examples ( 1 ) Is 79158435267 divisible by 229 ? Y 229x131—29999 P4= 3 / . t 791 5843 5267 ? I 5725 21644 \ But 5725=229X25 .% YES (2 ) Is 6056200566 divisible by 283 ? Y 283X53=14999 P3=15 6 056 200 566 6226 10414 8690 But 6226=283X22 .-. YES

( 294 ) (3) Is 7392 60251 divisible by 347 ? 347X317=109999 .\ P4= l l ( 7 3926 6251 ? (73654 0627 J But 73654=347X212 .\ YES (4) Is 867 311 7259 divisible by 359 ? V 359X39=14001 Q„=14 J8 11111

673 Tl7 764 3509

259 ? J

NO

(5) Is 885648437 divisible by 367 ? v 367X3=1101 j

8

~85

I 734

64 314

6 6

Q2= l l 84 323

37 ? J

But 734=367X2 .-.Y E S ( 6 ) Is 490 2 2 2 8096 divisible by 433 ? v 433X3=1299 .% Pa= 13 .-. ( 49 02 22 80 96 ? { 1257 1292 399 1328 J

NO.

( 7 ) Is 51 8 8 8 8 8 8 37 divisible by 467 ? v 467X3=1401 / . Qa= 14 ( 51 I —467

8 8

—37

8 8

8 8

504

430

37 ? J

YES

N .B. :— The alternative method of successive mechanical oscula tions is also, of course, available (but will prove generally less neat and tidy and will also be more tedious). (8 ) Is 789405 35994 divisible by 647 % v 647X17=10999 .\ P3= l l ( 78 940 535 994 ) j 1294 6110 11469 \ But 1294=647x2

YES

(9) Is 2093 1726 7051 0192 divisible by 991 ? V 991X111 = 110001 .-. Q4 = l l , .-. ( 2093 1726 7051 \ —30721 —526 —4939 But 30721=991X331 YES

0192 ? J

( 295 ) ( 1 0 ) Is 479466 54391 divisible by 421 ? v 421X19=7999 Ps = 8 J 47 946 654 391 ) I 1694 7205 3782 J NO ( 1 1 ) What change should be made in the first digit of the above number in order to render it divisible by 421 ? Answer:—As 1684 is exactly 4x421, the only change needed in order to reduce the actually present 1694 into 1684 is the alternation of the first digit from 4 to 3.

Ch a p t e r

XXXI

SUM AND DIFFERENCE OF SQUARES Not only with regard to questions arising in connection with and arising out of Pythagoras’ Theorem (which we shall shortly be taking up) but also in respect of matters relating to the three fundamental Trigonometrical-Ratio-relationships (as indicated by the three formulae Sin2 0 -fcos 2 0=1, 1 + ta n 2 0 = S ec2 0 and 1 -j-cot 2 0 = cosec2 0 ) etc., etc. we have often to deal with the difference of two square numbers, the addition of two square numbers etc., etc. And it is desirable to have the assistance of rules governing this subject and benefit by them. Difference of two Square numbers Of the two, this is much easier.

For, any number can

be expressed as the difference of two square numbers. The Algebraical principle involved is to be found in the elementary formula a2 -—b 2 = (a + b ) (a—b). This means that, if the given number can be expressed in the shape of the product of two numbers, our task is automatically finished. And this “ if” imposes a condition which is very easy to fulfil. For, even if the given number is a prime number, even then it can be correctly described as the product of itself and of unity. Thus 7 = 7 x 1 , 1 7 = 1 7 x 1 , 197=197x1 and so on. In the next place, we have the derived formula : (a + b ) 2 —(a—b) 2 = 4 a b ; and therefore ab can always come into the picture as / a-f-b \ 2 / a—b \2^ - e (half the sum)2—(half the diffe- I 2 / I 2 / rence)2 ; and as any number can be expressed as ab, the problem is readily solved. And the larger the number of factorisations possible, the better. In fact, if we accept fractions too as permissible, the number of possible solutions will be literally infinite.

( 297 ) For example, suppose we have to express 9 as the difference of two squares. We know that— 9 = 9 X 1 .\ 9 = ~ ‘ ' 2 = 5 2 —4 2 Similarly

(1) 1 3 x 1 = ( ¥ ) M ¥ ) 2 = 7 2 - 62

(2) 1 2 = 6 X 2 = 4 2— 22 or 4 x 3 = 3 | 2— |2 or 12 X l = 6 j 2— 5^2 (3) 4 8 = 8 X 6 = 7 2— l2 or 1 2 x 4 = 8 2— 42 or 16 X 3 = 9 f 2— 6|2 or 48 X 1 = 2 4 ^ 2— 23J2 or 24 X 2 = 1 3 2— ll2 The question, therefore, of expressing any number as the difference of two squares presents no difficulty at all ! The Sum of Two Square Numbers Inasmuch, however, as a 2 -f-b 2 has no such corresponding advantage or facilities etc., to offer, the problem of expressing any number as the sum of two square numbers is a tough one and needs very careful attention. This, therefore, we now proceed to deal with. A Simple Rule in Operation We first turn our attention to a certain simple rule at work in the world of numbers, in this respect. We need not go into the relevant original Sutras and explain them (especially to our non-Sanskrit-knowing readers). Suffice it for us, for our present purpose, to explain their purport and their application. Let us take a particular series of “ mixed” fractions, namely,— 1 J, 2 f, 3$, 4^ , 5T\ etc. which fulfil three conditions : (i) that the integer-portion consists of the natural numbers in order ; (ii) that the numerators are exactly the same * and 38

( 298 ) (iii) that the denominators are the odd numbers, in order, commencing from 3 and going right on. It will be observed that, when all these fractions are put into shape as “ improper” fractions, i.e.'as JJ M -44 ftS etc., etc. the sum of D 2 and N 2 is invariably equal to (N-f-1 )8 ! And the General Algebraical form being,

.-. D „=2n+1 ; and N=2n (n-fl) D2+ N 2= (N + 1 )* or. (2n + l)2+ 4 n 2(n + l)2=(2n2+ 2 | i+ l)a The shape of it is perhaps frightening; but the thing in itself is very simple : and the best formula is D2+ N 2= (N + 1 )2. This means that when a 2 (given)-(-x 2 is a perfect square, we can readily find out x 2. Thus, for instance, (i) If a (the given number) be 9, 2n+I = 9 n=4 .•. 4J is the fraction we want. And 92+ 4 0 2= 4 1 2 (ii) If a be 35, 2 n + l —35 n = 1 7 .\ The fraction wanted is 17$ | = 6,y 352+612*=6132. (iii) If a = 5 7 , 2 n + l= 5 7 .•. n = 2 8 The required fraction is 28|f=i#^-* 572+16242=1625a (iv) I f a=141, 2 n + 1=141 .-. n = 7 0 .-. The fraction wanted is 70xV> x=W t 141a+99402=99412 N ote:—Multiples and sub-multiples too behave in exactly the same manner (according to Anurupya i.e. pro portionately). For instance, Let a=35 .'. 2 n + l= 3 5 .\ n=17 .\ The fraction wanted is 17|| = 'r r .-. 702+12242=12262

352+612*=6132

A Simpler Method (for the same) This same result can also be achieved by a simpler and easier method which does not necessitate the “ mixed” fractions, the transforming of them into the “ improper” — fraction—shape etc., but gives us the answer immediately.

( 299 ) It will be observed that, in all the examples dealt with above, Since D2+ N 2= (N + 1 )2 D2= (N + 1 ) 2 —N 2 = 2 N + 1 = N + (H + 1 ) In other words, the square of the given number is the sum of two consecutive integers at the exact middle. For instance, if 7 be the given number, its square= 4 9 which can be split up into the two consecutive integers 24 and 25 .\ 72+ 2 4 2= 2 5 2. Similarly, ( 1 ) If a = 9 , its square (81)=40+41

.\ 92+ 4 0 2= 4 1 2

(2 ) If a=35, its square (1225)=612+613

.-. 352+612* =6132

(3 ) If a==57, its square (3249)=1624+1625 .\ 572+ 16242=16252 (4) If a=141, its square (19881)=9940+9941 .*. 1412+99402=^§9412 and so on. And all the answers are exactly as we obtained before (by the first method). The Case of Even Numbers Yes; the square of an odd number is necessarily odd and can be split up into two consecutive integers. But what about even numbers whose squares will always be even and cannot be split up into two consecutive numbers ? And the answer is that such cases should be divided off by 2 (and other powers of 2) until an odd number is reached and then the final result should be multiplied proportionately. For example, if a=52, we divide it by 4 and get the odd number 13. Its square (169)=84+85 .\ 132+ 8 4 2= 8 5 2 .\ (multiplying all the by 4), we say : 522+3362= 3402. There are many other simple and easy methods by which we can tackle the problem (of a2+ b 2= c 2) by means of clues and conclusions deducible from 3 2 + 4 2 = 5 2, 52+ I 2 2= 1 3 2, 8a+ 1 5 2 = 1 7 a etc. But into details of these and other allied matters we do not now enter.

Ch a p t e r

X X X II

ELEMENTARY SQUARING, CUBING ETC. In some of the earliest chapters of this treatise, we have dealt, at length, with multiplication-devices of various sorts, and squaring, cubing etc., are only a particular application thereof. This is why this subject too found an integral place of its own in those earlier chapters (on multiplication). And yet it so happens that the squaring, cubing etc., of numbers have a particular entity and individuality of their own ; and besides, they derive additional importance because of their intimate connection with the question of the square-root, the cube-root etc., (which we shall shortly be taking up). And, consequently, we shall now deal with this subject (of squaring, cubing etc.), mainly by way of Preliminary Revision and Recapi tulation on the one hand and also by way of presentation of some important new material too on the other. The Yavadunam Sutra (for Squaring) In the revision part of it, we may just formally remind the student of the YAVADUNAM formula and merely cite some examples thereof as a sort of practical memory-refresher : . 2 . 3. 4. 5. 6 . 7. 8 . 9. 1 0 . 1 1 . 1 2 . 13. 1

972= 9 4 /0 9 ; 872= 7 4 ,/169=7569 1 9 * = lls 8 8 l = 2 8 a 1=361 912=82/81 9652=9301/225=931 /225 113*=1261/69=12769 9962=992/016 9982=996/004 9997 2=9994/0009 1007 2= 1014/049 9996a=9992/0016 99r992=9998/0001 10172=1034/289

( 301 ) 14. 15. 16.

1039a=10781/521=1079/521 99991a==99982/00081 999982=99996/00004

17. 18.

999942=99988/00036 100042=10008/0016

19.

999978a==999956/000484

20. 21. 22.

999998a=999996/000004 1000232— 100046/00529 99998732=9999746/0016129

23.

9999999a=9999998/00000001

24.

10000122= 1000024/000144 The Anurupya Sutra (for Cubing)

This is new material. suffice to explain i t :

A simple example will, however,

Take the hypothetical case of one who knows only the cubes of the “ first ten natural numbers” (i.e. 1 to 1 0 ) and wishes to go therebeyond, with the help of an intelligent principle and procedure. And suppose he desires to begin with l l 3. . The first thing one has to do herefor is to put down the cube o f the first digit in a row (of 4 figures in a Geometrical Ratio in the exact proportion subsisting between them). Thus— 1

11 8= 1 1 1 1 2

2

13 3 1 (ii) The second step is to put down, tinder the second and third numbers, just two tunes the said numbers themselves and add up. And that is all ! A few more instances will clarify the procedure : (2) 13® = 1 3 9 27 (1) 12s = 1 2 4 8 18 6 4 8

16 2 8 1

17 2 8

19

8

7

2 2 2 19 7

( S0 2 (3) 14s

4 16 64 8 32

1

7

2

(5) 16*

1

4

4

2 1

s=

3 (6 ) 17* =

64 512 16 128 3

8

4 2 4

8

W 19s =

2

(1 0 )

2 2

*=

6

12 1 (13) 25* =

8

(15)

(1 2 )

7

6

6

5 9

8

8

8

2

8

8

8

4

6

2

8

4

(14) 32* = 27 18 12 36 24 32

5

8

16 32 64 32 64

13

20 50 125 40 100

15

3

9 81 729 18 162

1

10

1

12 18 27 24 36

8

1

16 16

II eo

( 1 1 ) 23* =

2

9

6

1

7 5

7 49 343 14 98

1

8

9

25 125 50

3

4

6

8

5

5

1

1 0

4

9

0

1

(9)

(4) 15s =

36 216 72

6 1 2

)

7

6

8

8

9s = ( 1 0 — 1 )* = 1000— 100+ 10—1 ? —200+20 \ =1000—300+30—1==729

(16) 97s

729 567 1134 912

6

441 882

343

7

3

or, better still, 97®=(100—3)* =1000000—30000+900 —27 —60000+1800 1000000—90000+2700—27 =912 6 7 3 N .B. :—If you start with the cube of the first digit and take the next three numbers (in the top row) in a Geometrical

( 303 ) Proportion (in the ratio of the original digits them selves) you find that the 4th figure (on the right end) is just the cube of the second digit! The Algebraical explanation hereof is very simple: If a and b are the two digits, then our chart reads: a *+ a 2 b + ab 2 + b 8 2 a 8 b + 2 aba a 3 + 3 a 2 b + 3a b 2 + b 8 and this is exactly (a + b ) 8 ! Almost every mathematical worker knows th is; but very few people apply i t ! This is the whole tragedy and the pathos of the situation! The Yavadunam Sutra (for Cubing) The same YAVADUNAM Sutra can, in view of the above, be applied for cubing too. The only difference is that we take here not the deficit or the surplus but exactly twice the deficit or the surplus as the case may be and make a few corresponding alternations in the other portions also, as follows: Suppose we wish to ascertain the cube of 104. Our base being 1 0 0 , the excess is 4. So we add not 4 (as we did in the squaring operation) but double that (i.e. 8 ) and thus have 104+8 ( = 112) as the left-hand-most portion of the cube. Thus we obtain 1 1 2 . Then we put down thrice the new excess multi plied by the original excess (i.e. 1 2 x 4 = 48) and put that down as the middle portion of the product.

112/48

And then we affix the cube of the original excess (i.e. 64) as the last portion thereof. And the 112/48/64 answer is complete. Some more illustrative instances are given below for familiarising the student with the new process (which is not really new but only a very useful practical application of the (a + b ) 3 formula just above described:

( 304 ) (1) 1033=109/27/27 (because 9 x 3 = 2 7 ; and 33=27) (2) 113s= 139/07/97 (because 39x13=507 and 13s=2197) 5 21 =1442897 (3) 10043=1012/048/064 (because 1 2 x 4 = 4 8 and 43= 6 4 (4) 100053=10015/0075/0125 (because 15 X 5=75 and 53=125 (5) 9963=988/048/064=988/047/936 (•.• 12 X 4= 48 and 43= 6 4 (6) 933=79/47 /iF(because 21 X —7=147 and —73= —343 1 —3 =804357 (7) 9991a=9973/0243/0729 (because —27X —9=243 and —93= —729=9973/0242/9271 (8) 100073=10021/0147/0343 (9) 99999s=99997/00003/0000f=99997/00002/99999 (10) 100012s=100036/00432/01728 (11) 999983=99994/00012/00008=99994/00011/99992 (12) 1000007s=1000021/000147/000343 (13) 9999923=999976/000192/000512 (because 2 4 x 8 =192 & 83=512 Fourth Power We know that (a + b )4= a 4+ 4 a 3b + 6 a 2b 2+4ab3+ b 4. This gives us the requisite clue for raising any given number to its fourth power. Thus, 11111 353 14 64 1 1 2 4 8 16 6 20 24 20

73

6

The Binomial Theorem The “ Binomial Theorem” is thus capable of practical application and—in its more comprehensive Vedic form—has thus been utilised, to splendid purpose, in the Vedic Sutras. And a huge lot of Calculus work (both Differential and Integral) has been (and can be) facilitated thereby. But these details, we shall hold over for a later stage.

Chapter X X X III STRAIGHT SQUARING Reverting to the subject of the squaring of numbers, the student need hardly be reminded that the methods expounded and explained in an early chapter and even in the previous chapter are applicable only to special cases and that a General formula capable of universal application is still due. And, as this is intimately connected with a procedure known as the Dwandwa Yoga (or the Duplex Combination Process) and as this is of still greater importance and utility at the next step on the ladder, namely, the easy and facile extrac tion of square roots, we now go on to a brief study of this pro cedure. The Dwandwa-Yoga (or the Du/plex Combination Process) The term “ Dwandwa Yoga” (or Duplex) is used in two different senses. The first one is by squaring ; and the second one is by Cross-multiplication. And, in the present context, it is used in both the senses (a2 and 2ab). In the case of a single (central) digit, the square (a2 etc.,) is meant; and in the case of a number of even digits (say, a and b equidistant from the two ends), double the cross-product (2ab) is meant. A few examples will elucidate the procedure. Denoting the Duplex with the symbol D, we have: ( 1 ) For

2

(2 ) (3 ) (4) (5 ) (6 )

7, D =49 34, D = 2 (12)=24 74, D = 2 (28)=56 409, D = 3 (3 6)+ 0= 72 071, D = 0 + 4 9 = 4 9

For For For For For

, D = 2 2= 4

(7) For 713, D =

2

(21)+12=43

( 8 ) For 734, D =

2

(2 8 )+ 3 2=65

39

( 306 ) (9) (1 0 ) (1 1 ) (12) (13) (14) (15) (16) (17) (18) (19)

D for 7346 = 2 X 42+ 2X 12= 108 D for 26734=16+36+49=101 D for 6 0 1 7 2 = 2 4 + 0 + l2=25 D for 7 32 1 5 = 7 0 + 6 + 4 = 8 0 D for 80607=112+0+36=148 D for 7 7 = 2 X 4 9 = 9 8 D for 521398=80+36+6=122 D for 746213=42+ 8+12=62 D for 12345679=18+28+36+20=102 D for 370415291=6+126+0+40+1=173 D for 432655897=56+54+32+60+25=227

This is merely a recapitulation of the Urdhva Tiryak process of multiplication as applied to squaring and needs no exposition. N ote:—If a number consists of n digits, its square must have 2n or 2n-l digits. So, in the following process, take extra dots to the left (one less than the number of digits in the given numbers). Examples( 1 ) 2072=40809 I =42849 2 4 \ }

. . 207 -----------4 0 809 2 4 4 2 849

(2) 2132= 4 4 1369=45369 (3) 2212=48841 (4) 3342= 9 8 3 4 6=11556 13 2 1

(5) 4252=

. .425 16/16/44/20/25=1806*25 ( 6 ) 5432=25 0 6 49=294849 4 4 2 (7) 8972 . . . 8 9 7 64/144/193/126/49=80409 Or 1 TO 32= 1 —2 + 1 —6 + 6 + 0 + 9 = 8 0 4 0 9 Or (by Yavadunam ^?m)=s784/1032^=80409

( 307 ) 8 8 9 (8 ) 8892=64/128/208/144/81 =790321

2 Or

1

1

1

1

=

0

0

0

1

1

1

1

1 -2 -9 + 0 + 3 + 2 + 1 ^ 7 9 0 3 2 1 Or (by Ydvadunam Sutra) 889a= 7 7 8 /l l l a= 7 89 / 321=790/321. /12 (9) 1113*= . 1

. - 1 1 1 3 2 3 8 7 6 9

(10) 2134^=4/4/13/22/17/24/16=4553956 (11) 32142=9/12/10/28/17/8/16=10329796 (12) 32472=9/12/28/58/46/56/49=10543009 (13) 67032= 36/84/49/36/42/0/9=44930209 (14) 31 -422=9/6/25/20/20/16/4=987-2164 (15)

07312= -0049/42/23/6/1= ‘ 00534361

(16) 89788= 64/144/193/254/193/112/64=80604484 _2

Or

1 1 0 2 2 = 1

/-2/l-5/0/4/4/4=80604484

Or (by Ydvadunam Sutra) 7956/10222=80604484 (17) 8887a=64/128/192/240/176/l 12/49=78978769

2 Or 11113=1-2-10-3+8769=78978769 Or (by Ydvadunam) 7774/1113a=7774/ 8769=78978769 /123 (18) 141.32a=l/8/18/14/29/22/l3/12/4=19971.3424 (19) 21345a=4/4/13/22/37/34/46/40/25=455609025 (20) 430312= 16/24/9/24/26/9/6/1 =185 X 66961 (2 1 ') 46325a= l 6/48/60/52/73/72/34/20/25=2146005625 (22) 73•214a=49/42/37/26/^0/28/l7/8/16=5360329796

Chapter X X X IV VARGAMtfLA (SQUARE ROOT) Armed with the recapitulation (in the last chapter) of the “ Straight Squaring method” and the practical application of the Dwandwayoga (Duplex Process) thereto, we now proceed to deal with the Vargamula (i.e. the Square Root) on the same kind of simple, easy and straight procedure as in the case of “ Straight Division” . Well-known First Principles The basic or fundamental rules governing the extraction of the square root, are as follows: (1) The given number is first arranged in two-digit groups (from right to left); and a single digit (if any) left over (at the left-hand-end) is counted as a simple group by itself. (2) The number of digits in the square root will be the same as the number of digit-groups in the given number itself (including a single digit (if any such there be). Thus 16 will count as one group, 144 as two groups and 1024 as two. (3) So, if the square root contains n digits, the square must consist of 2n or 2n-l digits. (4) And, conversely, if the given number has n digits, the square root will contain n or n+1 digits.

2

2

(5) But, in cases of pure decimals, the number of digits in the square is always double that in the square root. (6) The squares of the first nine natural numbers are 1 , 4 , 9 , 16, 25, 36, 49, 64 and 81. This means : (i) that an exact square cannot end in 2, 3, 7 or

8

;

(ii) (a) that a complete square ending in I must have either 1 or 9 (mutual complements from 10) as the last digit of its square r o o t;

( 309 ) (6 ) that a square can end in 4, only if the square root ends in 2 or 8 (complements); (c) that the ending of a square in 5 or 0 means that its square root too ends in 5 or 0 (respectively); (d) that a square ending in 6 must have 4 or 6 (complements) as the last digit in its square r o o t; and (e) that the termination of an exact square in 9 is possible, only if the square root ends in 3 or 7 (complements). In other words, this may be more briefly formulated thus: (a) that 1, 5, 6 and 0 at the end of a number reproduce themselves as the last digits in its square; (b) that squares of complements (from ten) have the same last digit. Thus, l 2 and 9 2 ; 2 2 and 8 * ; 3 2 and 7 2 ; 4 2 & 6 2 ; 5 2 and 5 2 ; and 02 and 102 have the same ending (namely 1,4,9,6 ,5 and 0 respectively); and (c) that 2,3,7 and 8 are out of court altogether, as the final digit of a perfect square. Readily Available First Data. Thus, before we begin the straight extracting of a square root by “ straight division” method, we start with previous knowledge of ( 1 ) the number of digits in the square root and (2 ) the first digit thereof. Thus— ( 1 ) 74562814 N = 8 N in square ro o t= N / 2 = 4 ; and the first digit thereof is 8 . %* (2 ) 963106713. N = 9 .\ N in the square r o o t= N -f 1 = 5 ; & the first digit thereof is 3. 2~ But

(3) (' 7104)2 must contain (4) (5) ^ 0064=-08 ( 6 ) V 000049 =007

8

decimal digits.

( 310 ) (7) V ‘ 0 0 0 0 7 ( 0 ) = , 0 ° 8 etc. ( 8 ) V '00000007= *0002 etc. (9) V

7^

'

3

But ( 1 0 ) y/^9=

V^0=

= '9

etc.

Modus operandi (of Straight Squaring) The procedure of Straight Squaring as inculcated in the Vedic Sutras is precisely the same as in Straight Division but with this difference, namely, that in the former the Divisor should be excatly double the first digit of the Square root. N .B.

As a single digit can never be more than 9, it follows therefore that, in our method of straight squaring, no divisor above 18 is necessary. We may, of course, voluntarily choose to deal with larger numbers ; but there is no need to do so. Initial Chart

We ^thus start our operation with an initial chart, like the samples given hereunder: 29 : 7 31 : 5 (1 ) 1 : 3 : 4 :

32 *

(3) 10

49 : 7 :

: 5

(5)

: 12

63 84 :

(®)

1

:

44

0 0 12

8

:

:

44 44 : :

96 : 4 : :

6

8

4 •

•

1

*

2

40

(4)

•

1 2

(9)

4 :

•

2

31 76 4

2

(8 )

61

13

14' •

6

12

: 7

6

73 16 •

9 8

60 : 84 : :

(1 0 )

0 0

6

4

2

2

0 0

( 311 ) (11)

10 :

73 69 42

(12) 18 :

90 : 61 71 74 : 9 9

Further Procedure Let us now take a concrete case (the extraction of the square root of, say, 119716) and deal with i t : (i) In the above given general chart, we have not only put down the single first digit of the square root wanted but also prefixed to the next dividend-digit, the remainder after our subtraction of the square of that first digit from the left-hand-most digit or digit-group of the given number. ^ 9 7 16* And we have also set down g . ! g • as our divisor, the exact d o u b l e -----------------------3 ■ * of the first digit of the quotient. * __________ ’ (ii) Our next Gross Dividend-unit is thus 29. Without subtracting anything from it, we simply divide the 29 by the divisor 6 and put down %\ • 9 7 1 6 : the second quotient-digit 4 and 6 : : 2 5 : the second remainder 5 in their "— . proper places as usual. ______________ ’ (iii) Thus our third Gross Dividend is 5 7 . From this we subtract 16 (the square of the second quotient-digit), ^ . 9 7 1 6 * get 41 as the Actual Dividend, divide it by 6 and set down the 6 : : 2 5 5 : Q (6 ) and the R (5) in their 7 # . proper places as usual. J------ !---------------(iv) Our third gross dividend-unit is 51. From this we subtract the Dwandwa Yoga (Duplex) (=48), obtain 3 as the remainder, divide it ^ by 6 and put down the Q 6 : : 2553 (0 ) and the R (3) in their 7 . 0Q rproper places. ----------------- -------

( 312 ) (v) This gives us 30 as our last gross dividend-onit. From this we subtract 36 (the Dwandwa Yoga of the third quotient digit This means

6

) ; get

that the work

that the given expression is a perfect square and that 346 is its square root. And that . „ H is all.

as Q and as R.

0

has been completed,

• 9 7 l 6 : 2 5 5 3 “ . . „ — : 3 : 4 6.0 0 — — ---------------11

6

:

COMPLETE

Proof of Completeness & Correctness ( 1 ) A manifest proof Of the Complete-squareness of the given expression (and of the correctness of the square root ascertained) is by squaring the latter and finding the square to be exactly the same as the given complete square. 3462—9/24/52/48/36=119716 (2 ) But this is too mechanical.

Thus,

YES. We obtain a neat and

valid proof from the very fact that, if and when the process is continued into the decimal part, all the quotient-digits (in the decimal part) are found to be zeroes and the remainders too are all zeroes ! Proof to the Contrary A number can not be an exact square in the following circumstances : (1) if it ends in 2, 3, 7 or

8

;

(2 ) if it terminates in an odd number of zeroes; (3 ) if its last digit is

6

but its penultimate digit is even ;

(4 ) if its last digit is not

6

but its penultimate digit is

o d d ; and ( 5 ) if, even though the number be even, its last two digits (taken together) are not divisible by 4 ; AND a square root cannot be correct if it falls to fulfil any of the requirements hereinabove indicated :

( 313 ) Examples Some instructive illustrative examples are given below : (1 )

4: :

5 :2 : 1

12

:

(2) 10

:

32: 4 9 : 7 4

9 6 4 1

: :

1 : 6 3 84 : 0 2 3 6

(4) 2

:

1

55 : 2 0 4 9 14:

6

6

: :

5 : 7 (complete:

: 4 (complete):

6

(5)

: :

: 3 (complete) :

2

40:

(3)

9 0

(«)

2

:

2

(complete):

8

: 69: 9 0 6 3 1 : 5 11 5 41 16:

7 : 4 3*00 (complete)

: 8 :3

6

1

.

0

0

0

A complete square (7) 14: 53:

1 6 3 2 1 4 4 13 6 13 5 7

7 : 2 9 1 *3 1...(incomplete) (8 )

6

: 14 : 0 4 5

8

7 5 0 4 11 13

7 4

3:

8

...(incomplete)

(9) 12: 41 : 2 5 4 9 2 9 5 4 5 2 1 0 6

(1 0 )

4:

1 0

:

: 25 : 5:

(1 2 )

1 2

: 45: 6

40

0 0 0 A A complete square

7: 3 8 9 1 5 4 3 5 5 13 6 7 4 2

(1 1 )

: 4 2 3

7 2

8

3

0 0 0 0

9

8

0 An exact square

7 4 5 4 7 6 0 7 4 5 5 1 0

7 4*000

. A perfect square

3 1 9 8 2 4 9 9 6 3 1

: 7 3 2 0 0 0

. A complete square

( 314 ) (13) 16

74

5 7 5 3 1 4 49 10 9 13 19 12 7 7 4

8

(14) 14

52

73 8

(16)

8

18 4

(17)

6

2 4 1 • 80

5 5*

75 8

(19)

6

0 0

0

(2 0 )

(2 1 )

8

10

3 3

6

2 *5

4

8

4

6

7 0 0 0

8 6

6

8

(to 4 decimal-places)

8

12

7 '24 24 0 1 7 1 0 0 0 0 (to 4 places) 11 14 9 11 10 6

09: :

1 24

6

2 4 0

2

3 :0

0 1 6

4*

0 9 7 2

(2 2 ) 16 •74 *8

(to 5 places)

3 9 7

790 0 0 0 7 7 14 19

27*

0

6

6

16*

5 *

(to 3 decimal-places)

134 5 1 2 6 (to 3 decimal-places) 2 5 9 10 16 17

4

(18) 16

/ . A n exact square

...

2 1 0 8 9 12 16 14 15

13 3

5 7*0 0 0

4 4 3 9 0 7 (to 2 places of decimals) 3 6 4 13 7

7 (15) 16

3

6

5

0

0

0

0

(to 5 places)

6

130 00 0 0 0 0 0 0 1 9 10 8 16 7

2 0

64

8

1 0 7 0 0 10 5 14 19 14 6

0

(to 5 places)

8

50

000 (to 5 places)

( 315 ) (23)

7 0

19-

8

: :

4(24)

3

4 3 9 1

27-

10

6 4 1 2 8 10 10 15 17 13

5

0

9 0 •

0 0

0

0

0

14 (to

6

places)

• 0

(to

0

6

places)

2 10 9 12 17 10 1 9

5(25)

6

0009 : :

0

1

1

2

:

0

2

2

2

03 (27)

1 3 4 0 0 0 0 0

(to

6

places)

•0039 : 3 0 0 0 0 0 0 0 (to eight places) : 3 9 14

12

: 2

06

6

8

9 7 1

00000083 : 1 0 0 0 0 0 0 0 (to : 2 3 11 18 7 2 0

(28) 18

•00 0 9 :

11

8

places)

5 9

000092 : 4 0 1 0 0 0 0 0 (to ten : 11 6 6 13 15 14 24 32

(29) 18

09

0

(30)

...

0 0 07 5

•3 (26)

2 4

0 0 4 5 13 (to 5 decimal-places) 0 0 0 4 3 1

09

6

1 5

6

2

:

6

1 2

5 4

38

0 7 3 6 12 3 1

2

:

1

: 4 4.

0

0

.•. A complete scfuare.

Or, taking the first two digits together at the first step, we have : 3 6 : 207 : 11

:

14:

1

0 4. 0

An exact square

Chapter X X X V CUBE ROOTS of EXACT CUBES (Mainly by Inspection and Argumentation) (Well-known) FIRST-PRINCIPLES ( 1 ) The lowest cubes (i.e. the cubes of the first nine natural numbers) are 1, 8 , 27, 64, 125, 216, 343, 512 and 729. (2) Thus, they all have their own distinct endings; and the is no possibility of over-lapping (or doubt as in the case of squares).

(3) Therefore, the last digit of the cube root of an exact cube is obvious: cube root ends in

(i) Cube ends in 1 ;

;

C R ends in 8

(iii) C ends in

3;

C R ends in 7

(iv) 0 ends in

4;

C R ends in 4

(v) C ends in

5;

C R ende in 5

(ii) C ends in

2

(vi) C ends in (vii) C ends in

6

8

;

6

C R ends in 3

7;

(viii) C ends in (ix) C ends in

C R ends in

;

1

C R ends in

;

2

; and

C R ends in 9

9;

(4) In other words, (i) 1, 4, 5, 6 , 9 and endings; and (ii) 2, 3, 7 and (from 1 0 ).

8

0

repeat themselves in the cube-

have an inter-play of complements

(5) The number of digits in a cube root is the same as the number of 3 -digit groups (in the original cube) including a single digit or a double-digit group (if any such there be).

( 317 ) ( 6 ) The first digit of the cube-root will always be obvious (from the first group in the cube). (7) Thus, the number of digits, the first digit and the last digit of the cube root of an exact cube are the data with which we start, when we enter on the work

of

extracting

the cube root of an exact cube.

Examples (Let a, 1 and n be the symbols for the first digit, the last digit and n the number of digits in the cube root of an exact cube). ( 1 ) For 271, 601, f = 6 , 1 = 1 and n = 2 (2 ) For 4, 269, 813, F = l , L = 7 and n = 3 (3) For 5, 678, f==l, L = 2 and n = 2 (4 ) For 33, 076, 161, F = 3 , L = 1 and n = 3 (5 ) For 83, 453, 453, F = 4 , L = 7 and n = 3 (6 ) For 105, 823, 817, f= 4 , L = 3 and n = 3 (7) For 248, 858, 189, f = 6 , L = 9 and n = 3 (8 ) For 1 , 548, 816, 893, f = l , L = 7 and n = 4 (9) For 73, 451, 930, 798, f= 4 , L = 2 and n = 4 ( 1 0 ) For 76, 928, 302, 277, f= 4 , L = 3 and n = 4 ( 1 1 ) For 6 , 700, 108, 456, 013, f = l , L = 7 and n = 5 ( 1 2 ) For 62, 741, 116, 007, 421, f= 3 , L = 1 and n = 5 (13) For 91, 0 1 0 , 0 0 0 , 0 0 0 , 468, f= 4 , L = 2 and n = 5 and so on. The Chart-Preliminary and Procedure The procedure is similar to the one adopted by us in “ Straight Division” and particularly in the extraction of square roots. The only difference is that our divisor (in this context) will not be double the first digit of the root but thrice the square thereof. As we know the first digit at the very outset, our chart begins functioning as usual (as follows) : (1)

3 : 1 : 7 :

2 8

: 0 1

1

(2) 12: 13 : :

: 5 2:

8 2 4

(3) 48 : 73 : 089149 : : 901 : (5)

4:

8

4

1

0

: 3:

12 : 21 ? 400 713 : : 13

: (7)

(4) 27 : 27 : : :

(6 ) 192 : 600 : :

132 419

: 88

2:

48 : 79 : 314 502 : : 15 :

4: Algebraical Principle Utilised

Any arithmetical number can be put into its proper algebraical shape as : a + 1 0 b + 1 0 0 c+

1 0 0 0

d etc.

Suppose we have to find the cube of a three-digit arith metical number. Algebraically, we have to expand ( a + 1 0 b + 1 0 0 c)3. Expanding it accordingly, we have : (a+10b+100c)8==a8-r 100b8+1000000c8+ 3 0a 2b+300ab2 +300a2c+30000ac2+300000b2c+300000bc2+6000abc. Removing the powers of ten and putting the result in algebraical form, we note the following: ( 1 ) The units’ place is determined by a8. (2 ) The tens’ place is determined by 3 a2b (3) The hundreds’ place is contributed to by 3 ab 2 + 3 p,2c (4) The thousands’ place is formed by b 8 + 6 abc (5) The ten thousands’ place is given by 3ac2 + 3 b 2c ( 6 ) The lakhs’ place is constituted of 3bc2 ; and (7) The millions’ place is formed by c3. N.B. :— The number of zeroes in the various coefficients (in the Algebraical expansion) will prove the correctness of this analysis.

( 319 ) Note :— If one wishes to proceed in the reverse direction, one may do §q ; and, for facility’s sake, the letters substi tuted (for a, b, c, d etc.) may be conveniently put down as L, K , J, H etc.

The Implications of the Principle This Analytical sorting of the various parts of the algebraical expansion into their respective places, gives us the necessary clue for eliminating letter after letter and determining the previous digit. And the whole procedure is really of an argumentational character. Thus, (i) From the units’ place, wTe subtract a3 (or L3) ; and that eliminates the last digit. (ii) From the ten’s place, we subtract 3a2b (or 3L 2 K) and thus eliminate the penultimate digit. (iii) From the hundreds’ place, we subtract 3a2 c+ 3 a b 2 (or 2 3L2J-|-31k2) and there-by eliminate the pre penultimate digit. (iv) From the thousands’ place, we deduct b3+6abc ; and so on

N.B. :—In the case of perfect cubes we have the additional advantage of knowing the last digit too, beforehand. Some instructive examples are given below : ( 1 ) Extract the cube root of the exact cube 33, 076, 161. Here a—3 ; L = 1 ; and n = 3 . (L) L = 1

.\ L 8= l .

1

(K) 3L 2 K = 3 k (ending in 6 ) Deducting 3K ; we have

1

1

I 33 076 161

.\ Subtracting 1, we have

K=

6

2

(J) 3L 2 J + 3 L K 2 = 3 J + 1 2 (ending in 3J ends in 9 .*. J = 3

33 0761 1

)

) i

CR=321

N.B. :— The last step is really unnecessary (as the first digit is known to us from the outset).

( 320 ) (2 ) Extract the cube root of the exact cube 1728. Here, a = l ; L —2 and n = 2 C R =12 (3) Extract the cube root of the exact cube 13,824 Here a = 2 ; L = 4 ; and n = = 2 / . CR=24 (4) Determine the cube root of the exact cube 83, 453, 4 5 3 . Here F = 4 ; L = 7 ; and N = 3 (L) L = 7 L 3 =343. ) 83 453 453 Subtracting this, we have r 343 J 83 453 1 1 (K) 3L2K==147K (ending in 1 ) K==3 l 441 subtracting 441 J 83 448 7 (J) 3L2J + 3 L K 2==147J-f-189 (ending in 7)1 147J ends in 8 J= -4 J -*• CR=437 N.B. :—Exactly as in the previous example. (5) Find out the cube root of the exact cube 84, 604, 519 Here a = 4 ; L = 9 ; and n = 3 84 604 519 (L) L = 9 L 3=729 Subtracting this ____ 222. (K) 3L*K=243K (ending in 9) .• .1 = 3 ) 846°729 Subtracting 729t 845965-----(J) 3L2J + 3 L K 2=243J+243 (ending in 5) i / . 243J ends in 2 J=4 1 CR=439 N.B. :—As before. (6 ) Extract the cube root of the exact cube Here a = 6 ; L = 9 ; and n = 3 .

2488 58189 2488 58189 729

(L) L = 9 L3=729 Subtracting this, 24885 746 (K) 3L2K=243K (ending in 6) i 486 K=2 Deducting 486) J 248852 6 (J) 3L2J+3L K 2=243J+108 (endingin 6) l 243J ends in

6

J=6)

J

CR=629.

N.B. Same as before. (7 ) Determine the cube root of the exact cube Here a = 4 ; L = 3 ; and n = 3 (L) L = 3 L3=27. Subtracting this

105823817 105823817 27

10582379 (K) 3L2K = 2 7K (ending in 9) Subtracting 189, we have

K =7

189 J 1058219

1

( 381 ) (J) 3LaJ + 3 L K 2= 2 7 J +441 (ending in 8 )\

J ••• C R = 473

J=4 N.B. As before, (8 ) Extract the cube root of the exact cube Here a = 5 ; L = 3 ; and n = 3 (L) L = 3 L8=27. deducting this (K) 3L2K = 2 7 K (ending in 4) .\JL= Subtracting 54, we have

2

143 055 667 143 055 667 ___________ 27 143 055 64 1 ________ 54 1143 055 1

(J) 3L2J + 3 L K 2= 2 7 J + 3 6 (ending in 1 ) \ J= 5 J

CR=523

N.B. Exactly as before.

(9) Find the cube root of the cube Here a = 4 , L = 3 ; and n = 4 .

76, 928, 302, 277. 1 The last 4 digis are 2277 J 27 225 (L) L = 3 .-. L®=27. Subtracting this, (K) 3L2K = 2 7 K (ending in 5) l 135 K =5 Subtracting 135x } 09 (J) 3L2J + 3 L K 2= 2 7 J +225 (ending \ in 9 ) .-. J = 2 J ••• CR=4253

N .B .:— But, if, on principle, we wish to determine the first digit by the same method of successive elimination of the digits, we shall have to make use of another alge braical expansion (namely, of ( L + K + J + H ) 3. And, on analysing its parts as before into the units, the tens, the hundreds etc., we shall find that the 4th step will reveal 3L2H + 6 L K J + K S as the portion tc be deducted. So, (H)

L 2 H + 6 L K J + K 8= 2 7H + 1 80 +1 25 = 2 7 H + 3 05 (ending in 3) .-. H = 4 : and CR=4253

3

i 30 : 09 2 : 7» 1 27 : 3

( 1 0 ) Determine the cube root of the cube 11,345, 123, 223 Here a = 2 ; L = 7 ; and n = 4 11 345 123 223 (L) L = 7 / . L 3 =343. Subtracting i t , ___________ 343 11 345 12 288 41

( 322 ) (K) 3L2K +147K (ending in 8 ) K = 4 . Deducting 588 (J) 3LJ2+ 3 L K a=147J+336 (ending in 0 ) .\ J —2 Subtracting 330, we have (H) 3L2 H + 6 L K J + K 3=147H +336 + 6 4 = 1 4 7 H -f400 (ending in 4). H= 2 N.B.

-i 588 J ll 345 1 1 70 •)________ 330 Cl 134508 4 ) -j [ CE=2247 )

The last part (for ascertaining the first digit) is really superfluous.

( 1 1 ) Extract the cube root of the cube Here a = 2 ; L = 7 ; and n = 4 . (L) L = 7 .’. Ls=343 Deducting it

1 2 1 2

12

, 278, 428, 278 428

4 4 3 4 4 3

343 278 428 10

(K) 3L2K =147K (ending in 0 ) i 0 K=0. J1 2 278 428 1 (J) 3L2J + 3 L K 2= 1 4 7 J -f0 (endingin l)i 441 J=3 J 1 2 278 384 Subtracting 441, we have (H) 3L2H + 6 L K J + K * = 1 4 7 H + 0 + 0 1 and ends in 4 H= 2 J CR=2307 N.B.

As in the last example.

( 1 2 ) Find the cube root of the cube Here a = 7 ; L = 1 ; n = 4 (L) L = 1 L 3 = l . Deducting it, (K) 3L 2 K = 3 K (ending in 4 ) K= 8 Deducting 24.

355 045 312 441 355045 312 441 _______ 1 355045312 44 24

(J) 3L 2 J + 3 L K 2= 3 J + 1 9 2 (ending in 2 ) J= 0 (H) 3L2H + 6 L K J + K 3= 3 h + 0 + 5 1 2 and ends in 3 h= 7

192 355045293 CR=7081

N.B. Exactly as above. Note:— The above method is adapted mainly for odd cubes. If the cube be even, ambiguous values may arise at each step and tend to confuse the student’s mind.

( 323 ) (13) Determine the cube root of the cube 792 994219216 Here a = 9 ; L = 6 ; and n = 4 792. 994 219 216 (L) L = 6 A L 3=216* Deducting this, ____________ 216 792 994 219 00 (K) 3L2K =108K (ending in zero) ________5^0 A K = 0 or 5. Which isdt to be ? 772 994 2136 Let us take 5 (a pure gam ble)! (J) 3L2J + 3 L K 2= 108J+450 (ending 1 7729961 in 6 ) / . J — 2 or 7 ! Which should we prefer ? Let us accept 2 (another perfect gamble) I . (H) 3L2H + 6 L K J + K 3-1 0 8 H -f 360 +125=108H +485(ending in 7) .% H = 4 or 9 ! Which should we choose ? Let us gamble again and pitch for 9 !

6 6 6

47

A CR==9256

Here, however, our previous knowledge of the first digit may come to our rescue and assure us of its being 9 . But the other two were pure gambles and would mean 2 x 2 i.e. four different possibilities! A Better Method At every step, however, the ambiguity can be removed by proper and cogent argumentation ; and this may also prove interesting. And anything intellectual may be welcomed ; but it should not become too stiff and abstract; and An ambiguity (in such a matter) is wholly undesirable (to put it mildly). A better method is therefore necessary, is available and is given below. All that has to be done is to go on dividing by 8 (until an odd cube emanates), work the sum out and multiply by the proper multiplier thereafter. Thus, 8 ) 792 994 249 216 8

) 99124 281152

8

) 12390535144.

1 548 816 893 1

1

1

( 824 ) Here a = l ; L==7 ; and n = 4 (L) L = 7 a L 8 =343. Subtracting this, f ) (K) 3L3K =147K (ending in 5) ^ £ •\ K = 5 . Deducting 735. )

1548816893 3 4 3

154881655 7 3 5

15488092 672

(J) 3L2J + 3 L K 2=147J+525 (ending in \ 2 ) /. J = 1 Deducting 672, we have : J1548 742 (H) 3L2H + 6 L K J + K 8=147K-)-210 +125=147H +335 (ending in 2 ) /. H= 1 and CR (of the original cu b e= 8 N.B.

*\ The cube* (root is 1157 JAnd X 1157=9256

Here too, the last step is .unnecessary (as the first digit is already known to us).

(14) Determine the cube rot of the cube 2 , 840, 362, 499, 528 Here a = l ; L = 2 ; and n = 5 (L) L = 2 L8 = 8 ^2840 362 499 528 8

Deducting this, we have (K) 3L 2 K = 1 2 K and ends in 2 K = 1 or 6 ! Let us take Deducting 1 2 K.

J2840 362 499 52 72 6

! (2840 362 4988 ) __________ 228

(J) 3L2J3LK2=12J+216 (ending }2840 362 476 in 8 ) J = 1 or 6 ! Let us take 1 ! (H) 3L2 H + L K J + K 3 = 1 2 H + 1 2 ^ + 216=2H +238 (ending in 6 ) ( / . H = 4 or 9 : Let us take 4 ! ) (G) We need not bother ourselves about G and the expansion of ( a - f b + c + d + e ) 3 and so on. Obviously G = 1

CR=14162

But the middle three digits have been the subject of un certainty (with 2 x 2 x 2 = 8 different possibilities). We

( $25 ) must therefore work this case too out by the other—the unambiguous—method. or

8

: 2 8 4 0 3 6 2 4 9 9 5 2 8 3 5 5, 0 4 5, 3 1 2 , 4 4 1

Here a = 7 ; L =1 ; and n = 4 (L) L = 1 ,\L3= 1 } Subtracting this, we haver )

355 045 312 441 1 355 045 312 44

(K) 3L 2 K = 3 K (ending in 4) ^ K= 8 Subtracting 24, J

24 355 045 312 2

(J) 3L2J + 3 L K 2= 3 J + 1 9 2 -n and ends in 2 /. J = 0 C Subtracting 192, )

19 2 355 045 13

(H) 3L2H + 6 L K J + K 3= 3 H + 0 + 5 1 2 and ends in 3 /. H = 7 .\ Cube Root=7081 CR of the original expression—14162 (15) Find out the 6741.

1 2

-digit exact cube whose last four digits are

Here a = ? ; L = (L) L =

1

1

; and n = 4

.\ L 3= l

. .

(K) 3L 2 K = 3 K and ends in 4 K = 8 / . Deducting 24.

1

i J

(J) 3L2J + 3 L K 2= 3 J + 1 9 2 and ends \ in 5 J=1 J /. Subtracting 195, we have (H) 3L 2 H + 6 L K J + K 3= 3 H + 4 8 + 512=3H +560 and ends in 7. H =9 N.B.

7 4 1

. 6

Subtracting it, . . .

6

. . .

6

7 4 2 4 5

. . 1 9 5 7

^ , .\ The original J) Cube Root is 9181

As we did not know the first digit beforehand, all the steps Were really necessary.

(16) A 13-digit perfect cube begins with 5 and ends with 0541. Find it and its cube root.

( 326 ) Here a = l ; L = 1 ; and n = 5 . (L) L = 1 .'. L 3 = l . Deducting it.

. . .0 5 4 1 1

0 5 4 2 4 . . . 0 3 2 13 . . . 9

(K) 3L2 K = 3 K K= 8 i Subtracting 24, we have J (J) 3L2J + 3 L K 2= 3 J + 1 9 2 and ends in 3 3=1 { Deducting 213, we have (H) 3L2H + 6 L K J + K 8= 3 H + 3 3 6 ) + 5 1 2 = 3 H + 8 4 8 and ends in 9 ) H =7 .-. CR= 17781 (J) And Gr=l And the cube= 17781s

Chapter X X X V I CUBE ROOTS (GENERAL) Having explained an interesting method by which the cube roots of exact cubes can be extracted, we now proceed to deal with cubes in general (i.e. whether exact cubes or not). As all numbers cannot be perfect cubes, it stands to reason that there should be a general provision made for all cases. This, of course, there i s ; and this we now take up. First Principles It goes without saying that all the basic principles ex plained and utilised in the previous chapter should hold good here too. We need not, therefore, re-iterate all that portion of the last chapter but may just, by way of recapitulation, remind ourselves of the conclusions arrived at there and the modus operandi in question. The Sequence of the Various Digits (1 ) (2 ) (3) (4) (5) (6 ) (7)

The The The The The The The

first place by a3 second place by 3a 2b third place by 3ab2 + 3 a 8c fourth place by 6 a b c-fb 3 fifth place by 3ac2 + 3 b 2c sixth place by 3bc2 seventh place by c 3 ; and so on.

The Dividends, Qotients, and Remainders (1 ) (2 ) (3) (4) (5) (6 ) (7 )

The first D, Q and R are available at sight. From the second dividend, no deduction is to be made. From the third, subtract 3ab2 From the fourth, deduct 6 a b c + b 8 from the fifth, subtract 3ac2 + 3 b 2c from the sixth, deduct 3bc2 from the seventh, subtract c 8. ; and so on.

( 328 ) Modus Operandi Let us take a concrete example, namely, 258 474 853 and see the modus operandi actually at work (step by step)

258 : 4 74853 108 : : 42 100 :

(а) Put down

6

6

: 3

and 42 as first Q and first E by mere

VUokanam (inspection). (б ) The second Gross Divident is thus 424. Don’t subtract any thing thereform. Merely divide it by 108 and put down 3 and

1 0 0

(c) So, the third Gorss Dividend is 1007. Subtract therefrom 3ab* (i.e. 3 X 6 X 3 2 i.e. 162). The

: 258 : 4 • •^ 2 ; 6 : 3

74853 1 0 0 7

as Qa and Ra

1 ^ 8

: 258 : 4 : : 42

7

4853 89

1 0 0

: 6 : 37 ---------- -----------------------

third Actual Working Dividend therefore is 1007— 162=845. Divide this by 108 and set dovra 7 and 89 as Qs and R8 (d) Thus, the fourth Gross 258 : 4 Dividend is 894. Sub- *08 : : 42

7 1 0 0

4 89

853 111

tract therefrom 6 abc+ : 6 : 37 0 c3= (7 5 6 + 2 7 =783). ------------ ------------------------So, the fourth actual working dividend is 894—783= 111. Divide this again by 108 and put down 111 as Q4and R4.

0 and

(e) Our next gross dividend is 258: 4 now 1118. Subtract there- 108 : : 42

7 4 8 5 3 100 89 111 47 6- 3 7 0 0 0

from 3ac2+ 3 b 2c=882+189 . =1071. Therefore our fifth ----- '________________ ~ actual working dividend is 4 7 . Divide it by 108 and put down 0 and 47 as Qg and R 5

( 329 ) (/) Our sixth gross dividend is 4 7 5 . Subtract therefrom 3bc2(=441) So, our Qfi and R ft 6 6 now are 0 and 34.

*^ 8

258: 4 7 4 8 5 3 : :42/l 00/89/111/47/34:

6

:

3 7 0 0 i , “T T (complete cube)

(g) Our last gross dividend is thus 3 4 3 . (=343) therefrom and set down 0 and and 6 7

0

0

Subtract C8 as our Q 7

This means that the given number is a perfect cube, that the work (of extracting its cube root) is over and that the cube root is 637. N.B.

Proof of the correctness of our answer is, of course, readily available in the shape of the fact that 6378 is the given number. But this will be too crudely and cruelly laborious. Sufficient proof, however, is afforded by the very fact that, on going into the decimal part of the answers, we find that all the quotients and all the remain der sare zeroes. An Incomplete cube is now dealt with as a sample : Extract the cube root of 417 to 3 places of decimals Here the divisor is 147. 417* : 0 0 0 0 147 : 3 43 : 74 152 155 163

(a) Here Qx and R j= 7 and 74 (b) .\ The second gross dividend is 740. No subtraction is required. Dividing 740 by 147, we get 4 and 152 as Q 2 and R 2. (c)

The third gross dividend is 1520. Subtracting 3ab2 (=336) therefrom, we have 1184 as our third actual working dividend. We divide it by 147 and put down

7

and 155 as our Qa and R 3.

(d) Our fourth gross dividend is 1550. We subtract 6 a b c+ b 3 (=1176+64=1240) therefrom, obtain 310 as our 42

( 330 )

fourth actual working dividend, divide it by 147 and set down 1 and 163 as our Q4 and R 4. (e) Our next gross dividend is 1630. We subtract 3ac2+ 3b2c (=1029+336=1365) therefrom, get 265 as our fifth actual working dividend, divide it by 147 ; and so on. N ote:— The divisor should not be too small. Its ultra-smallness will give rise to big quotients (sometimes of several digits), the insufficiency of the remainders for the subtractions to be made and other such complications which will confuse the student’s mind. In case the divisor actually happens to be too small, two simple devices are available for surmounting this difficulty. (i) Take the first four (or 5 or 6 ) digits as one group and extract the cube root. For example, suppose we have to find out the cube root of 1346, 085. Our chart will then have to be framed thus: 363 : 1 , 346 : 085 : 1 331 : 15 :

:

11

Let us now take an actual concrete example and apply this method for extracting the cube root of 6334625 : : 6334 : 6 2 5 972 : 5 832 : 502 166 312 1944 :

18

:

5

0

2

........

(а) Q1=18 ; R ,=502 ; and Divisor (D)=972. (б ) No subtraction being needed at this point, divide 5026 by 972 and put down 5 and 166 as Q2 and R2. (c) Our third gross dividend (G.D) is 1662 ; subtract 3ab2 (=1350) from 1662, divide the resultant. Actual dividend (AD) i.e. 312 by 972 and set down 0 and 312 as Q9 and R8.

( 331 ) (d) Our next GD is 3125. Subtract 6abc+b3 (=0+125^= 125) from 3125, divide the AD (3000) by 972 and put down 2 and 1944 as Q4 and R4 and so on, Or, Secondly, multiply E (the given expression) by 2 3, 3 3, 4 3 or 5 3 etc., (as found necessary and most convenient) find the cube root and then divide the CR by 2 , 3 , 4 , 5 etc. For instance, instead of taking 3 (as the divisor), take 3 X 43 (= 3 X 64=192), find the cube root and divide it by 4. Here again, a concrete example may be worked out by both the methods: First Method (а) Qr= l ; R x= l ; and D = 3

2

3 : : (б ) Now, G D = A D = 1 0 .

1

: 0 0 0 0 : 1 4 10 2 : 26

Divided by 3, it gives

2

and 4 as

Q2 and R>2* (c) The third GD is 40. From this subtract 3ab2 ( = 1 2 ). After this subtraction, the AD is 28. Divide this by 3 and put down 6 and 1 0 as Qa and R s. (d) The fourth GD is 1 0 0 . From this deduct 6abc+b8 (= 7 2 + 8 = 8 0 ). The AD is 2 0 . Now, as for dividing this 2 0 by 3, the directly apparent Q4 and R4 are 6 and 2 . Rut the actual quotient and Remainder are difficult to determine (because of the smallness of the divisor) and the insufficiency of the Remainders for the next subtrac tions and a good number of trial digits may fail before one can arrive at the correct figures! This is why the other method is to be preferred in such cases. And then the working will be as follows: Multiplying 2 by 53, 250 : 0 0 0 we get 250. 108 : : 34 124 196 332 : 6 : 2 9 9 ... (a) Qx and R x = 6 and 34 (b) E 2=340. Dividing this by 108, we have Q 2 = 2 and R 2=124.

( 332 ) (c) 3ab2=72. Deducting this from 1240, we get llftg, Dividing this by 108, Q8= 9 and R 8= l9 6 (d) 6 abc+ bs= 6 4 8 + 8 = 6 5 6 .*. The Working Dividend = 1960—656=1304. Dividing this by 108, we have Q4 = 9 and R 4=332 .*. The CR=6*299— Dividing by 5, the actual cube root=l*259— ( 2 ) Let us take another concrete example i.e.

1500 : 0 ^63 :______ : 169

0

0

238

400

387'

. We multiply : n • 4 4 7 ~ 12 by 5 3 and put 1500 down as the Total Dividend. And we take the first four digits as one group. 3

1 2

(а) Thus Q1= l l and R x=169 (б ) Dividing 1690 by 363, we have Q 2 =

and R 2=238

4

(c) 3ab2=528 .*. Working Dividend=2380—528=1852 Dividing it by 363, we have Q8 = 4 and R =400 (d) 6 a b c+ b 3=1056+64=1120. Deducting this from 4000, we get 2928. Dividing this by 363, Q4= 7 and R =387 .*. The C R = 1 1*447 etc. .*. The cube root of the original E =2*289. .. Some more examples may be taken : ( 1 ) (a) E=1728 ; Qx= l ; D = 3 ; and R 2 = 0

3 : :

1 : 7 2 8 : 0 1 0 1

:

2

*0

0

(exact cube) (6 )

7

divided by 3 gives

2

and

1

as Q 2 and R 2

(c) Third Gorss Dividend= 1 2 ; 3ab2 = 1 2 ; dividend= 0 .*. Q8= 0 and R3= 0

.*. Actual

(d) Fourth gross dividend=8; 6abc+b3= 0 + 8 = 8 .*. Subtracting the latter from the former, Q4= 0 and R4= 0 The CR=12

^

( 333 ) N.B.

The obvious second proof speaks for itself.

(2 ) (a) Here E=13824 ; D = Qt= 2 ; and R 1 = 5

1 2

; 12

1 3 : 8 2 4 : :5 10 6 : 2 : 4 0 0 (Perfect cube)

(b) 51 gives Qa= 4 ; and R a =

1 0

(c) Gross Dividend=102; 3ab®=96. Actual dividend= 6 . Divided by Q3 =

0

and R 8 =

1 2

, this gives

6

(d) G .D = 6 4 ; 6 abc+ b3= 0 + 6 4 and R4==0

A.D =

0 Q4 = 0 The CR is 24.

( 3 ) Here E =33, 076, 161 ; Qx= 3 ; D =27 ; and R ^ (a) Q, = 3 ; and R , = 6 33 : 0 7 6 1 6 1 27 : : 6 6 4 2 0 0 : 3 : 2 10 0 0 (complete cube). (6 ) G D = A D = 6 0 ; Divided by 27, this gives

2

& 6

Q2 and Ra

as »

(c) GD is now 67 ; 3ab2= 3 6 A.D =31. 27, this gives us 1 and 4 as Qa and Rs.

Divided by

(d) GD is 46; 6 abc+ b3= 3 6 + 8 = 4 4 .‘ . A D = 2 And, divided by 27, this gives 0 and 2 as Q4 and R4. (e) GD is

21

; 3ac2 + 3 b 2 c = 9 + 1 2 = 2 1

divided by 27, this gives

0

and

0

A D =0 ; and, as Q5 and R 6.

(/) GD is 6 ; 3 bc 2 = 6 AD=0. And, divided by %1, this gives us 0 and 0 as Q6 and Re. (g) GD = 1 ; c3= l ; A D = 0 . 0 and 0 as Q 7 and R 7

Divided by 27, this gives us .-. The CR is 321.

N.B.— The second proof is clearly there before us. (4 ) E = 101

101 : 0 0 0 0 48 : : 37 82 148 112 40 (а) Q j= 4 ; R != 3 7 ; and D =48 : 4. : 6 5 9 5

(б ) G D = A D =370 ; and, divided by 48, this gives us and 82 as Qa and R 2

6

( 334 ) (c) GD=820 ; 3ab8=432 .\ AD=388. and, divided by 48, this gives us 5 and 148 as Qs and R s (d) G.D=1480 ; and 6 ab c+ b 8= 720+216=936 ; AD=544. And this, divided by 48, gives us and 1 1 2 as Q4 and R 4

9

(e) GD = 1 1 2 0 ; and 3ac2+ 3 b 2c =300+540=840 AD=280. And, divided by 48, this gives us and 40 as Q6 and R 6 ; and so on.

5

(5) E=29791 (а) Here C^—3 ; and R x=

2

; and D =27

(б ) G D = A D = 2 7 ; and, divided by 27, this gives us 1 and 0 as Qa and R a.

29 : 7 9 1 27 : : 2 0 : 3: 1 .0 (complete cube)

(c) G D = 9 ; and 3ab2= 9 ; A D = 0 , and, divided by 27, this gives us 0 and 0 as Q , and R a The CR is 31. N .B. :— The proof is there-before us as usual. ( 6 ) The given expression (E )=83, 453, 453 (a) Qx= 4 ; R x= 19 ; and D ==48. 4 5 3 4 5 3 83 : (b) G D = AD=194. : 19 50 61 82 47 34 48: divided And, 4 : 3 7. 0 0 0 (exactcube) 48, this by gives us 3 and 50 as q 2 and r 8 (c) GD=505 ; and 3ab2=108 AD=397. And, divided by 48, this gives us 7 and 61 as Q8 and R # (d) GD=613 ; and 6 abc+ b8=504 + 2 7 = 53 1 A D =82 And, divided by 48, this gives us 0 and 82 as Q4 & R 4 (e) GD=824 ; and 3ac2+ 3 b 2c= 588+ 189= 777 AD =47 And this, divided by 48, gives us 0 and 47 as Q6 & R j (/) GD—3bc2= 4 75 —441 = 34. (g) GD=343 ; and c8=343 The CR is 437.

Q „= o and R e = AD =

0

Q7 = R

3 4

7= 0

< 838 ) N .B .:—The proof is there as usual. (7) E = 8 4 , 604, 519 (а) Q t = 4 ; j)= 4 8

84:

.

48 :

and R 1= 2 0

:

6 : 20

4:

0 62

4 80

5 129

3 9 •0

1 80

9 72

0 (perfect cube)

0

(б ) G D =A D =206. And, divided by 48, this gives us 3 and 62 as Q2 and R 2 (c) GD=620 ; and 3ab2=108 .-. AD=512. And, divided by 48, this gives us 9 and 80 as Qs and R 3 (d) GD=804 ; and 6abc+b®=648-f27=675 AD=129. And, divided by 48, this gives us 0 and 129 as Q4 & R 4 (e) G D = 1295; .-. AD =80.

and 3aca+3b*c=972+243=1215 Aild, divided by 48, this gives us 0 and

80 as Qg and (/) GD=801 ; and 3bca=729 A D =72. And, divided by 48, this gives us 0 and 72 as Q 8 and R # (g) GD=729; and C8=729 .-. AD = .•. The CR is 439

0

.-. Q 7 =

0

and R 7 =

0

N .B .:— The proof is there as usual. ( 8 ) E=105, 823, 817 (а) Q 1 =

4

R i = 4 1

; .

and D = 4 8

: 105: 8 2 3 8 1 7 48 : : 41 82 90 56 19 2 :

4

;

7 3'

0

0

0

(complete cube)

(б ) G D =A D =418. And, divided by 48, this gives us 7 and 82 as Q2 and R a (c) GD=822 ; and 3ab2=588 .-. AD =234 ; and, divided by 48, this gives us 3 and 90 as Qs and R s (d) GD=903 ; and 6 a b c + b 8= 504+343=847 .-. AD =56. And, divided by 48, this gives us zero and 56 as Q4 & R 4 (e) G D = 568; and 3ac2+ 3 b 2c= 108+ 441= 549 .-.AD =19. And divided by 48, this gives us zero and 19 as Qg & R j

( 336 ) (/) GD=191 ; and 3bc2=189 .*. AD = 2 ; and, divided by 48, this gives us zero and 2 as Qe and R 6 (g) GD=27 ; and C3=27 The CR=473

AI) =

0

.*. 0 7= 0 and R 7 =

0

N.B. :— The proof is there as usual. (9 ) E=143, 055, 667

143 : 0 5 5 6 6 7 75 :____ : 18 30 20 17 5 2 : 5 : 2 3 * 0 0 0 (exact cube)

(a) Qx = 5 ; R x—18 ; and D =75 (b) G D = A D =180 ; and, divided by 75, this gives us 2 and 30 as Q2 and R 2 (c) GD=305 ; and 3ab2= 6 0 AD==245 ; and, divided by 7 5 , this gives us 3 and 2 0 as Q3 and R a (d) GD=205 ; and 6 ab c+ b 3= 1 8 0 + 8 = 1 8 8 .*. AD =17. And, divided by 75, this gives us 0 arid 17 as Q4 and R 4 (e) GD=176 ; and 3ac2+ 3 b 2c= 1 3 5+ 36 = 1 71 .*. A D =5. And, divided by 75, this gives 0 and 5 as Q5 and R 5 (/) G D =56 ; and 3bc2= 5 4 .*. A D = 2 ; and, divided by 75, this gives 0 and 2 as Q6 and R 6. (g) GD=27 ; and c3=27 .*. A D = 0 Q? = 0 and R 7 = 0 /. The CR is 523 N.B. :— The proof is there as usual. ( 1 0 ) E = 2 4 8 ,858, 189.

(a)

: 248 : 8 5 8 1 8 9 108 : : 32 112 81 162 155 72 : 6 : 2 9 . 0 0 0 (perfect cube)

; ^=32;

and D=108

(b) G D =A D =328. And, divided by 108, this gives us 2 and 1 1 2 as Q 2 and R 2 (c) GD=1125 ; and 3ab2= 72 .*. AD=1053 ; and, divided by 108, this gives us 9 and 81 as Qs and R 3. (d) GD=818 ; and 6 ab c+ b 3= 6 4 8 + 8 = 6 5 6 .*. AD = 162. And, divided by 108, this gives 0 and 162 as Q4 & R 4 (6) GD=1621 ; and 3ac2+ 3 b 2c=1458-hl08=1506 A D = 5 5 . And, divided by 108, this gives us 0 and 5 5 as Q5 and R 5.

( 337 ) (/) GD=558 ; and 3bc2=486 .\ A D = 7 2 ; and, divided by 108, this gives us 0 and 72 and Q6 and .Re (g) GD=729 ; and C3=729 .\ AD = 0 ; Q 7 = 0 and R 7 = 0 The CR is 629 N .B. ;—The proof is there as usual. (11) 11, 345, 123, 223 N ote:—The cube root in this case being of four digits, the method obtained from the expansion of (a + b - f c ) 3 will naturally not suffice for this purpose ; and we shall have to expand ( a + b - f c - f d ) 3 and vary the above pro cedure in accordance therewith. This is, of course, per fectly reasonable. The Schedule of Digits The Analytical digit-schedule for (a+b-f-c-|-d ) 3 now stands as follow s: (а) First digit ( 9 zeros)=a3— ( б ) Second digit ( 8 zeros)=3a 2 b — (c) Third digit (7 zeros)=3ab 2 + 3 a 2c (d) Fourth digit (6 zeros)= 6 a b c + b 3+ 3 a 2d (e) Fifth digit (5 zeros)= 6 a b d + 3 a c 2 + 3 b 2c — (/) Sixth digit (4 ze ro s )= 6 a cd + 3 b c2+ 3 b 2c — (g) Seventh digit (3 z e r o s )* = 6 b c d + 3 a d 2+ c 3— (h) Eighth digit ( 2 zeros)= 3 bd2+ 3 c 2d — (i) Ninth digit

(1

zero)=3cd2—

(j) Tenth digit (no zero)= d 3 Consequent Subtractions ( 1 ) Qj and R x by mere inspection. (2 ) Q 2 and R 2 by simple division (without any subtraction whatsoever). ( 3 ) From all the other Gross Dividends, subtract: (3) 3ab2 (4) 6 a b c + b 3 (5) 6abd+3ac 2 + 3 b 2c ( 6 ) 6acd+3bc 2 + 3 b 2d

( 338 )

(7) 6bcd+3ad*+c 8 ( 8 ) 3bda+3c*d (9) 3cda (1 0 ) d8 respectively, in order to obtain the actual working dividend and thence deduce the required Q and R. Note :—-It will be noted that, just as the equating of d to zero in (a -f-b -fc+ d ) 3 will automatically give us ( a + b + c ) 8 exactly so will the substitution of zero for d in the above schedule give us the necessary schedule for the three-digit cube root. As we go higher and higher up (with the number of digits in the cube root), the same process will be found at work. In other words, there is a general fbrmula for n (n being any positive integer); and all these are only special applications of that formula (with n equal to 2, 3, 4 and so on). This general form of the formula, we shall take up and explain at a later stage in the student’ s progress. In the meantime, just now, we explain the application of the (a -f-b + c + d ) 3 schedule to the present ease. Application to the Present Case 12 :

1 1 : 3 4 5 1 2 3 2 2 3 : 3 9 2 2 37 59 76 69 62 34

: 2 : 2 4 7' 0 (а) Qx=

2

; R1=

3

(б ) G D = A D = 33. R a= 9

0

0

; and Divisor is

0 (exact cube) 1 2

Dividing this by

(c) G D =94 ; and 3ab2=24.

1 2

A D =70

, we get Q a =

2

and

Q3= 4 and R s =

(<*) GD==225 ; and 6 abc+ b3= 9 6 + 8 = 1 0 4 ; Q4= 7 and R 4=37

AD =

2 2

1 2 1

(e) GD=371 ; and 6abd+3ac24-3b2c= 1 6 8 + 9 6 + 4 8 = 3 1 2 . AD =59 .-. Q 5 = 0 and R s=59. (/) GD—592; and 6acd+3bc2+ 3 b 2d = 3 3 6 + 9 6 + 8 4 = 5 1 6 D =76 Q# = 0 and R #=76.

( 339 ) (g) GD=763 ; A D =69

and

6bcd+3ad2+ c 8==336+294+64=694

v Q7 —

0

and R 7=69

(h) G D = 6 9 2 ; and 3bd2-+3c2d = 294+ 336= 630 Q8 = 0 and R 8= 6 2 (i) GD=622 ; and 3cd2=588 R *=34

A D =34

(j) GD=343 ; and d3=343 .\ A D = The cube root is 2247

0

AD =62.

.V Q0=O and

.*. Q10= 0 and R10= 0

N.B. :—The ocular proof is there, as usual. This is the usual procedure. There are certain devices, however, which can help us to over-come all such difficulties ; and, if and when a simple device is available and can serve our purpose, it is desirable for us to adopt it and minimise the mere mechanical labour involved and not resort to the other ultra-laborious method. The devices are therefore explained hereunder : The First Device The first device is one which we have already made use of, namely, the reckoning up of the first 4, 5 or 6 digits as a group by itself. Thus, in this particular case : : 11, 345 : 1 2 3 2 2 3 1452 : 10 648 : 697 1163 412 363 62 34 :

22

:

4

7.

0

0

0 (complete cube)

(a) Qx (by the same process) is the double-digit number 2 2 ; R 1=697 ; and D=1452 (b) G D =AD =6971 ; and, divided by 1452, this gives us 4 and 1163 as Qa and R2 (c) G D =11632; and 3ab2==1056 .\ AD=10576 ; and, divided by 1452, this gives us 7 and 412 as Qs and R 8 (d) GD=4123 ; and 6ab c+ b 3=3696+64=3760 AD= 363 ; and, divided by 1452, this gives us 0 and 363 as Q4 and R 4

( 340 ) (e) G D = 3632; and 3ac2+ 3 b 2c=3234+336=3570 .-. A D =62 ; and, divided by the same divisor 1452, this gives us zero and 62 as Q6 and R 5 (/) G D = 6 2 2 ; and 3bc2=588 .\ AD =34.

So Q6 =

0

and R 7 =

0

and R 6—34 (g) GD=343 ; and c3=343 .*. AD = The C.R.=2247 N.B. (

0

Q7 =

0

1 ) And the Proof is there before us, as usual, (2 ) By this device, we avoid the complication caused by shifting from (a + b -fc ) 3 to ( a + b + c + d ) 3. It, however, suffers from the draw-back that we have first to find the double-digit Q, cube it and subtract it from the first five-digit portion of the dividend and that all the four operations are of big numbers.

Second Device This is one in which we do not magnify the first group of digits but substitute (c+ d ) for c all through and thus have the same ( a + b + c ) 3 procedure available to us. But, after all, it is only a slight alteration of the first device, whereby, instead of a two-digit quotient-item at the commencement, we will be having exactly the same thing at the end. The real desideratum is a formula which is applicable not only to two-digit, three-digit, or four-digit cube roots but one which will be automatically and universally applicable. But this we shall go into at a later stage of the student’s progress. In the meantime, a few more illustrative instances are given hereunder (for further elucidation of—or at least, the student’s practice in, the methods hereinabove explained : ( 1 ) E = 1 2 , 278, 428, 443. Here too we may follow the full procedure or first ascertain the first two-digit portion of the cube root of 12, 278, treat the whole five-digit group as one packet and extract the cube root of the whole given expression in the usual way. The procedure will then be as follows :

( 341 ) (i) Single-digit method 1 2 : 2 7 8 4 2 8 4 4 3 12 : : 4 6 13 27 22 33 44 3 34 : 2:

307

0

0

(а) Q j= 2 ; R x= 4 ; and D = (б ) G D = A D = 4 2

00 1 2

Qa= 3 and R a =

(c) GD=67 ; and 3ab2= 5 4 ; R s=13 (d) GD=138 ;

(perfect cube)

6

AD =13

Q3 =

and 6abc+b3= 0 + 27=27

and

0

AD =

1 1 1

Q4= 7 and R 4=27 (e) GD=274 ; and 6abd+3ac8+ 3 b 2c= 2 5 2 + 0 + 0 = 2 5 2 ; AI ) = 2 2 Q5 = 0 and R 5 = 2 2 (/) GD = 2 2 2 ; and 6 acd + 3bc2 + 3b2d = AD =33 Q 6 = 0 and R 6= 33 (g) G D = 3 3 8 ; AD= 4 4

0

+

0

+ 189

and 6bcd+3ad2+ c 8= 0 + 2 9 4 + 0 = 2 9 4 Q7 = 0 and R 7= 44

(h) G D = 444; and 3bd2+ 3 c 2d= 4 4 1 + 0 = 4 4 1 Q8 = 0 and R 8= 3 (i) G D =34 ; and 3cd2= 0 R 9= 34

A D =34

(j) GD=343 ; and d»=343 and R 1 0 = 0 The CR=2307

Q9 =

(ii) Two-Digit method : 12

12

: 278

:

:

:

2:

4

3...

Q,j (of two digits) is 23 ; 12278 : 4 2 8 4 1587 : 12167 : 111 1114 33 338 23

: 0

0

and

A D = A 0 , Q1 0 =

N.B- —The proof is before us, as usual.

Preliminary Work

AD=3

7 . 0

0

( 342 ) (а) Qj^ (of two digits)=23 ; R 1= l l l ; and D=1587 (б ) G D =A D =1114

Q2 =

0

; and R 2=1114

(c) GD =11142; and 3ab2= 0 .'. AD=11142 R a=33 (d) GD=338 ; and 6 a b c + b 3 = R =338

Q3==7 &

AD =338 .\ Q4 =

0

(e) GD=3384 ; and 3ac2+ 3 b 2c=3381 +0=3381 Q5 = 0 and R 6= 3 (/) G D =34 ; and 3bc2= 0

A D =34

Q6 =

(g) GD=343 ; and d3=343 A D = 0 , Q7 = The CR is 2307

0

0

0

&

AD =3 & R 6= 34

and R 7 =

0

N.B. :— The proof is before us, as usual. (2) E = 76, 928, 302, 277. (i) Single-digit method : 76 : 9 2 8 3 0 2 2 7 7 48 : : 12 33 44 56 59 44 29 13 2 : (а) Qx= 4 ; R x =

4: 1 2

2 5

3.

0

0

0

0

(Exact cube)

; and D = 48

(б ) G D =A D =129 .-. Q 2 =

2

(c) GD=332 ; and 3ab2= 4 8 R 3= 4 4

and R 2= 33 AD =284 .-. Q8 =

(d) GD=448 ; & 6 ab c+ b a= 2 4 0 + 8 = 2 4 8 Q4= 3 and R 4= 56

AD =

5

&

2 0 0

(e) GD=563 ; & 6abd+3ac2+ 3 b 2c= 1 4 4 + 3 0 0 + 6 0 = 5 0 4 .-. AD = 5 9 .-. Q6 = 0 and R 5=59 (/) GD=590 ; & 6acd+3bc*+3b2d = 3 6 0 + 1 5 0+ 3 6 = 5 46 .•. A D = 4 4 ; .•. Qe = 0 and R e= 4 4 (g) G D =442 ; and 6bcd+3ad2+ c s= 180+ 108+ 125= 413 .-. A D =29 ; .-. Q7 = 0 and R 7= 29 (h) GD=292 ; and 3bdz+ 3 c 2d = 5 4 + 2 2 5 = 2 7 9 .\ A D = 1 3 Qs= 0 and R 8=13

( 843 ) (i) G D = 1 3 7 ; and 3cd2=135

AD =

Q „ = 0 and

2

R j= 2 (j) G D = 2 7 ; and ds= 2 7 .-. A D = 0 R10= 0 The CR=4253

. . Q1 0 = 0

and

N.B. :—Proof as usual. (ii) Double-digit method or, secondly, .•. Qx (double-digit)= 4 2

: 76 : 48: :

12

: 4:

9 33

2

: 769 28: 3 0 2 2 7 7 5292 : 740 8 8 : 2840 1943 404 137 13 2 : 42

:

5

3

0. 0 0

(а) Qx= 4 2 ; R =2840 ; and D=5292 (б ) GD= AD=28403

Qa= 5 ; and R 2=1943

(c) GD=19430 ; and 3ab2=3150 Q*=3 and R „=404 (d) G D =4042; / . AD=137

and Q4 =

(c) G D =1372; A D =13

and .’. Q6 =

(/) G D = 1 3 7 ;

and

0

.’ . AD=16280

6abc+bs=3780+125=3905 aiid R 4=137 3ac2+ 3 b 2c=1134+225=1359 and R 6=13

0

3bca=135

and R #= 2 (g) G D =27 ; and c8=27

AD =

AD =

0

2

N.B. Exactly as above. (3) E =355, 045, 312, 441 (i) Single-Digit method. 355 :

0

:

1 2

7

:

(а) Qx= 7 ; R x= (б ) G D = A D = 1

0

8

1 3 8

3 1 2 4 4 1 39 55 19 6 0

1 0

; and D = 147

1 2 2 0

5

4

120 28

;

Q2 =

0

0

; Q7= 0 and R 7= 0

*'• The CR is 4253

147 :

Q6 =

and R g =

1 2 0

( 344 ) (c) G D = 12 0 4 ; & R s= 28

and 3aba= 0

(d) G D=285 ; and 6 a b c + b 3 = R 4=138

AD=1204

Qs =

8

AD =285 ; Q4= l and

0

(e) GD=1383 and 6abd+3ac2+ 3 b 2c= 0 + 1 3 4 4 + 0 = 1 3 4 4 A D =39 Q 5 = 0 and R 5=39 (/) GD=391 ; A D =55 (g) G D = 55 2 ;

and Q6 = and

A D =19 (A) G D = 19 4 ;

0

6acd+3bc2+ 3 b 2d = 3 3 6 + 0 + 0 and R 6= 55

6bcd+3ad2+ c 8= 0 + 2 1 +512=533

Q7 =

and R 7= 19

0

and

3bd2+ 3 c 2d = 0 + 1 9 2

AD =2

Qg=0 and R 8= 2 (i) G D = 2 4 ; and 3cda= 2 4 (j) GD =

A D = 0 ; Q = 0 and R = 0

; and d8= l

1

A D = 0 , Q=

and R =

0

0

The CR is 7081

N.B. As above. (ii) Double-digit method. : 14700 :

355045 : 3 1 2 343000 : 12045 2853 391

: (a) Qx (of

70 2

:

1 * 0

8

4 4 1 40 2 0 0

0

digits)= 7 0 ; R x= 12045 ; and D=14700

(,b) G D =AD =120453

Q2 =

8

and R 2=2853

(c) GD=28531 ; and 3ab2=13440 Q3= l and R s=391

.-.

AD=15091

{d) GD=3912; & 6 a b c+ b 3=3360+512=3872 Q4= 0 and R 4= 4 0

A D =40

(e) GD=404 ; & 3ac2+ 3 b 2c= 2 1 0+ 19 2 = 40 2 Q5 —O and R 5= 2

AD =2

(/) G D =24 ; and 3bc2= 2 4

AD =

0

.-. Q#= 0 and

R #= 0

(g) G D = 1 ; and d3= l AD= 0 The CR is 7081

Q7= 0 and R 7= 0

( 345 ) N.B. As above. (4) E=792, 994, 249, 216 (i) Single-Digit method. 243 :

792: 9 9 4 2 4 9 2 1 6 : 63 153 216 158 199 152 72 56 21 9

5

6

.0

0 00

(а) Qx= 9 ; R x=63 ; and D =243 (б ) G D = A D =639

Q2 =

2

; and R a=153

(c) GD=1539 ; and 3ab2=108 and R 3=216

.\ AD=1431

(d) GD=2164 ; & 6 ab c+ b s= 5 4 0 + 8 = 5 4 8 Q4 = 6 and R 4=158 (e) GD=1582 ; and 6 abd + 60=1363 AD =199 (/) GD=1994 ; 72=1842

3

Qs

AD=1616

ac2 + 3b*c = 648 + 675 + Q5= 0 and R B=199

and 6acd+3bc2+ 3 b*d=1620-f 150-jAD =152 Q „ = 0 and R 6=152

(g) GD=1529; & 6bcd+3ad2+ c 3=360+972+125=1427 A D = 72 / . Q7= 0 and R 7=72. (h) GD=722 ; & 3bd2+ 3 c 2d = 216+ 450= 666 A D =56 Qg= 0 and R 8= 56 AD=21 \ Q9= 0 a n d (i) GD=561 ; and cd2=540 R#= 2 1 (j) GD=216 ; and ds=216

AD =0

Q10= 0 and

R io = ° The CR=9256 N.B. As above. Double-Digit Method 243 : :

792 : 9 : : 63 153 : 9 : 2 :

792994 25392 : 778688 92 44

digits is 92 and And D=25392.

Q2 (of two 923=778688. 9 0

2

0 0 0 0

1

6

( 346 ) (а) Q i=92 ; and R 1=14306 (б ) G D = A D = 143062 Qa= 5 ; and R 2=16102 (c) G D =I61024; and 3ab2=6900 AD=I54124 Q8 = 6 ; and R 8=1772 (d) GD=17729 ; and 6abc+b8= 16560+125=16685 AD =1044 Q4 = 0 and R 4=1044 (e) G D = 10442 ; and 3ac2+ 3 b ac=9936+450=10386 A D =56 .\ Q5 = 0 and R 5=56 (/) GD=561 ; and 3bc2=540.\AD=21.\Q6= 0 ; & R e = 2 (g) GD=216 ; and D8=216 / . AD = 0 Q7 = 0 and R 7 = The CR is 9256

1 0

N.B. As above. N ote:— It must be itted that, although the double-digit method uses the ( a + b + c ) 8 schedule and avoids the ( a + b + c + d ) 8 one, yet it necessitates the division, multiplication and subtraction of big numbers and is therefore likely to cause more mistakes* It is obviousby better and safer to use the ( a + b + c + d ) 3 and deal with smaller numbers. In this particular case, however, as the given number terminates in an even number and is manifestly divisible by 8 (and perhaps 64 pr even 512), we can (in this case) utilise a third method which has already been explained (in the imme diately preceding chapter), namely, divide out by 8 (and its powers) and thus diminish the magnitude of the given number. We now briefly remind the student of that method. Third Method : 7 9 2 8

8

99 4

249

216

9 9

124

281

152

:

390

535

144

1 2

1

4

5

8

5

7

816,

8

16

6

2

1

548,

,

0

1

893 6

36

55

0

0

8

74 0

1

( M7 ) (a) Q ,= l and R 2 = (6 ) G D = A D = 0 5

0

Qa= l and R a =

(c) 6 D = 2 4 ; and 3ab®=3

AD =

2

Q8= 5 and R 8 =

2 1

(d) G 0 = 6 8 ; and 6 ab c+ b 8= 3 0 + l = 3 l Q4 = 7 ; and R 4= 1 6

6

A D =37

(e) GD=168 ; and 6 ab d + 3ac*+ 3b«c= 42+ 75+ 15= 132 A D =36 Q6 = 0 and R 5= 36 (/) GD=361 ;an d 6a cd + 3b c2+ 3 b 2d = 2 1 0 + 7 5 + 2 1 = 3 0 6 A D =55 Q6 = 0 and R 6=55 (g) GD=556 ; and 6bcd+3ad2+ c 8=210+147+125= 482 A D =74 Q7 = 0 and R 7= 74 (h) GD=748; and 3bd2+ 3 c 2d = 147+ 525= 672 .\ A D =76 Q8 = 0 and R 8= 7 6 (i) GD=769 ; and 3cda=735 .\ A D =34 Q0=O ; and R #= 34 (j) GD=343 ; and d8=343 R io —

AD =

Qi0 =

0

0

; and

0

The C.R. (of the sub-multiple)=1157 The C.R. of the given number=9256 Or, Fourthly, the derived submultiple may be dealt with (by the two-digit method) thus :— 1548 : 8 1 6 8 9 3 363 : 1331 : 217 363 265 221 76 3 4 :

11

:

5

7

0

0

0

.(a) Qx= l l ; R x=217 ; and D =363 (b) G D =A D =2178

Q2= 5 and R a=363

(c) GD=3631 ; and 3aba=825 .\ AD=2806 .\ Q8 = 7 & R 8=265 (d) GD=2656 ; and 6 abc+ b8= 2 3 1 0 + 125=2435 AD= 2 21 Q4 = 0 and R 4 = 2 2 1 (e) GD=2218 ; and 3aca+3b*c=1617+525=2142 A D =76 Q5 = 0 and R e= 7 6

( 348 ) (/) GD=769 ; and 3bc2=735

A D =34

Q6 =

0

; &

R e=34 (g) G D=343; and c3=343 xAD=0 .\ Q7= 0 and R7= 0 .*. The cube root of the sub-multiple is 1157 .\ The CR o f the original number=9256 N.B. As above. (5 )

E= 8

: :

2

,

840,

: 147 : :

362,

355,

499,

045,

312,

528 441,

355 : 0 4 : 12 120 7

:

0

8

1

This very number having already been dealt with (in example 3 of this very series, in this very portion of this subject), we need not work it all out again. Suffice it to say that, because, 7081 is the cube root of this derived sub-multiple, The C.R. of the original number is 14162 Note :—All these methods, however, fall in one way or another, short of the Vedic ideal of ease and simplicity. And the general formula which is simultaneously applicable to all cases and free from all flaws is yet ahead. But these matters we shall go into, later.

Ch a p t e r

X XX V II

PYTHAGORAS5 THEOREM ETC. Modern Historical Research has revealed—and all the modern historians of mathematics have placed on record the historical fact that the so-called “ Pythogoras’ Theorem” was known to the ancient Indians long long before the time of Pythagoras and that, just as although the Arabs introduced the Indian system of numerals into the Western world and distinct ly spoke of them as the “Hindu” numerals, yet, the European importers thereof undiscerningly dubbed them as the Arabic numerals and they are still described everywhere under that designation, similarly exactly it has happened that, although Pythagoras introduced his theorem to the Western mathematical and scientific world long long afterwards, yet that Theorem continues to be known as Pythagoras’ Theorem ! This theorem is constantly in requisition in a vast lot of practical mathematical work and is acknowledged by all as practically the real foundational pre-requisite for Higher Geo metry (including Solid Geometry), Trigonometry (both plane and Spherical), Analytical Conics, Calculus (Differential and Integral) and various other branches of mathematics (Pure and Applied). Yet, the proof of such a basically important and fundamental theorem (as presented, straight from the earliest sources known to the scientific world, by Euclid etc., and as still expounded by the most eminent modern geometricians all the world over) is iiltranotorious for its tedious length, its clumsy cumbrousness etc., and for the time and toil entailed on i t ! There are several Vedic proofs, everyone of which is much simpler than Euclids’ etc. A few of them are shown below :

( 300 ) First Proof Here, the square A E =the square KG the four congruent right-angled traingles all around it. Their areas are c 2, (b—a) 2 and 4 x fab respectively. C2 = a 2 - 2 a b + b 2 -f4(| ab )= a 2 - f b 2 Q.E.D. Second Proof C o n s t r u c t io n :

C D = A B = m ; and D E = B C = n . .-. ABC and CDE are Congruent; and ACE is rightangled Isosceles. Now, the trapezium A B D E =A B C +C D E +A C E £mn+|h2+ lm n = i(m + n ) X (m4-n)==im 2 4-m n -fin 2 .-. 4h2 = $ m 2 -H n 2 .-. A2 = m 2 + n 2 Q.E.D. (N.B.

Here we have utilised the proposition that the area of a trapezium= | the altitude X the sum of the parallel Third Proof Here, A E = B F = C G = D H = m and E B = •FC=G D =H A= n. Now, the squae AC=the square E G + the 4 congruent right-angled triangles around it .•. h 2 + 4 (^ m n )= (m + n ) 2 = m 2 -|-2mn-fii2 .% A2 = m 2 - f » 2 Q.E.D.

Fourth Proof (The proposition to be used here is that the areas of similar triangles are proportional to the squares on the homo logous sides). Here, BD is _]_ to AC .-. The triangles ABC, ABD and BCD are similar. As between ( 1 ) the first two triangles and (2 ) the first and third ones,

( 351 )

M!=ADB. and^ B C D AC*

ABC ;

AC2 ABC

AB2-fBC2=A C 2 Q.E.D. Fifth Proof (This proof is from Co-ordinate Geometry. And, as modern Conics and Co-ordinate Geometry (and even Trigono metry) take their genesis from Pythagoras’ Theorem, this process would be objectionable to the modern mathematician. But, as the Vedic Sutras establish their Conics and Co-ordinate Geometry (and even their Calculus), at a very early stage, on the basis of first principles and not from Pythagoras’ Theorem (sic), no such objection can hold good in this case. The proposition is the one which gives us the distance between two points whose co-ordinates have been given. Let the points be A and B and let their co-orditiates be (a, 0) and (0, b) respectively. Then,

BA

V (a- ° ) 2+ ( 0 - b ) 2=

V a 2+ b 2 .-. BA2= a 2+ b 2 Q.E.D. N ote:—The Apollonius’ Theorem, Ptolemy’s Theorem and a vast lot of other Theorems are similarly easy to solve with the aid of the Vedic Sutras. We shall not, however, go into an elaborate description thereof (except of the Apollonius Theorem) just now but shall reserve them for a higher stage in the student’s studies.

C h a p te r X X X V I I I

APOLLONIUS’ THEOREM Apollonius5 Theorem (sic) is practically a direct and elementary corollary or offshoot from Pythagoras5 Theorem. But, unfortunately, its proof too has been beset with the usual flaw of irksome and needless length and laboriousness. The usual proof is well-known and need not be reiterated here. We need only point out the Vedic method and leave it to the discerning reader to do all the contrasting for himself. And, after all, that is the best way. Isn’t it. % Well, in any triangle ABC, if D be the mid-point of BC, then AB2+ A C 2= 2(AD2+ B D 2). This is the proposition which goes by the name of Apollonius5 Theorem and has now to be proved by us by a far simpler and easier method than the one employed by him. ly.

Let AO be the perpendicular from A

on BC ; let XO X ’ and YOY’ be the axes of co-ordinates ; and let BO, OD and OA be m, n and p respectively

D B = D C = m -fn A B a4-AC2= ( p a+ m 2) + ( m 2+ 4 m n + 4 n 2- f p 2) = 2 p 2 + 2 m 2 -f 4 m n - f 4 n 2

and 2 (AD2+ B D 2)= 2 [ (p2+ n 2)+(m 2-f 2mn-fn2) ] ~ = 2 p 2+2m 2+4m n+4n2 AB2+A C 2= 2 (AD2+ D B 2) Q.E.D.

( 363 ) the circum-centre, the in-centre, the ex-centree, the centroid, the Nine-Points-circle etc., can all be similarlyproved (very simply and very easily) by means of Co-ordinate Geometry. We shall go into details of these theorems and. their Vedic proofs later on ; but just now we would just merely point out that, like the “Arabic numerals” and “Pythogoras’ Theorem” , the “ Cartesian” co-ordinates are a historical M is n o m e b , no more, and no less.

Chapter X X X IX ANALYTICAL CONICS Analytical Conics is a very important branch of mathema tical study and has a direct bearing on practical work in various branches of mathematics. It is in the fitness of things, there fore, that Analytical Conics should find an important and pre dominating position for itself in the Vedic system of mathe matics (as it actually does). A few instances (relating to certain very necessary and very important points connected with Analytical Conics) are therefore given here under (merely by way, let it be remem bered, of illustration). I.

Equation to the Straight Line,

For finding the Equation of the straight line ing through two points (whose co-ordinates are given. Say, (9, 17) and (7,-2). The current method tells us to work as follows: Take the general equation y==mx+c. Substituting the above values therein, We have:

9 m + c= 1 7 ; and

7

m + c = —2 .

Solving this simultaneous equation in m and c, we have ; 9 m + c= 1 7 7m -f c = —2 2m =19 .*. m =9| Substituting this value of m (in either of the above two equations) we have, 0 6 | + c = — 2 Substituting these values of m and c in the Original General Equation (y=m x-f-c), we get y==9j X —6 8 j. .*. Removing fractions, we have 2 y= 1 9x— 137. And then, by transposition, we say, 19x—2y=137. But this method is decidedly too long and cumbrous (and especially for such a petty matter)!

( 355 ) And the Second Current Method (which uses the formula _ _ y 2 —y x / v _ v v is equally cumbrous and confusing. y~~Yi— i* It ultimately amounts to the right thing ; but it does not make it clear and requires several more steps of .working! But the Vedic at-sight, one-line, mental m ethod (by the Pardvartya Sutra) enables us to write the answer mechanically down by a mere casual look at the given co-ordinates. And it is as follows: The General Equation to the straight line (in its final form) is .. x— .. y = . . [where the co-efficients of x and y (on the left hand side) and the independent (on the Right hand-side) have to be filled in]. The Sutra tells us to do this very simply by : (i) putting the difference of the y —co-ordinates as the x-coefficient and vice versa ; and (ii) evaluating the independent term on that basis. For example, in the above example, the co-ordinates are : (9, 17) and (7, —2). (i) so our x-coefficient is 17—(—2)=19 (ii) and our y —coefficient is 9—7 = 2. Thus we have 19 X — 2 y as our L.H.S. straightaway. (iii) As for the abs6 lute term on the R.H.S., as the straight line in question es through the two given points, the substitution of the original co-ordinates of each of the points must give us the independent term. So, the substitution of the values 9 and 17 in the L.H.S. of the equation gives us 1 9 x 9 —2 x 1 7 = 1 7 1 — 34=137 ! Or Substituting the values 7 and — 2 therein, we get 19x7 —2x—2 = 1 3 3 + 4 = 1 3 7 ! And that is additional confirmation and verification ! But this is not all. There is also a third method by which we can obtain the independent term (on the R.H.S).

( 866 ) And this is with the help of the rule about Adyam Antyam and Madhyam i.e. be—ad (i.e. the product of the means minus the product of the extremes)! So, we have 1 7x7—9X — 2= 119+ 18= 137 ! And this is still further additional confirmation and verification ! So, the equation is :— 19x-—2y=137 which is exactly the same as the one obtained by the elaborate current method (with its simultaneous equations transpositions and substitu tions etc; galore) ! And all the work involved in the Vedic method has been purely mental, short, simple and easy ! A few more instances are given below: ( 1 ) Points (9, 7) and (—7, 2 ) A The Equation tothe straight line ing them is : 5x—1 6 y = —67 (2 ) (3) (4) (5) (6 ) (7) (8 ) (9)

( 1 0 , 5) and (18, 9) .\ x = 2 y (by Vilokan too) ( 1 0 , 8 ) and (9, 7) x —y = 2 (by Vilokan too) (4, 7) and (3, 5) 2x—y = l (9, 7) and (5, 2) A 5 x -4 y = 1 7 (9, 7) and (4, —6 ) /. 13x—5y=82 (17, 9) and (13, - 8 ) 17x—4y=253 (15, 16) and (9, —3) /. 19x—6y=189 (a, b) and (c, d) x(b—d)—y (a—c)=bc—ad II.

The General Equation and Two Straight Lines.

The question frequently arises:—When does the General Equation to a straight line represent two straight lines ? Say, 12x2+ 7 x y —10y2+ 1 3 x + 4 5 y —35=0. Expounding the current conventional method, Prof. S. L. Loney (the world-reputed present-day authority on the subject) devotes about 15 lines (not of argument or of explanation but of hard solid working) in section 119, example 1 on page 97 of his “ Elements of Co-ordinate Geometry” , to his model solution of this problem as follows : Here a = 1 2 , h— I, b = — 1 0 , g = ~ > f= ^ a n d c = Z Z Z

— 35

( $57 ) abc+2fgh—af2—bg2—ch2 = 1 2 (x — 10) ( x - 3 5 ) + 2 X ^ x H x | - 1 2 ^ j 2

- ( - ' » ) ( y ) ' - < - 56> ( I ) ’

= 4200+ ^ - 6075+ — + ^ = - 1875+ 1^ 2=0

4 4 4 The equation represents two straight lines.

4

Solving it for x, we have : x 2 + x 7 y + 13.|_^2 y + 13y _ 10y2- 4 5 y + 35+ ^ 7y + 1 3 ^ _ / 2 3 y —43v2 V 24 / . t _l 7 y + 1 3 •• + 24

= 2i i z i 3 24

... x = ? £ = ? or 1 ?1 ± ^ 3 4 .•.The two straight lines are 3 x = 2 y —7 and 4 x = —5y+ 5 jVote ;—The only comment possible for us to make hereon is that the very magnitude of the numbers involved in the fractions, their multiplications, subtractions etc., ad infinitum is appalling and panic-striking and that it is such asinine burden-bearing labour that is responsible for, not as a justification for, but, at any rate, an ex tenuation for the inveterate hatred which many youngsters and youngstresses develop for mathematics as such and for their mathematics-teachers as such ! We make no reflection on Prof. Loney. He is perhaps one of the best, the finest and the most painstaking of mathe maticians and is very highly esteemed by us as such and for his beautiful publications (which are standard authorities on the various subjects which they deal with). It is the system that we are blaming, (or, at any rate, comparing and contrasting with the Vedic system). Now, the Vedic method herein is one by which we can immediately apply the “ tJrdhwa” Sutra the Adyam Adyena

( 358 ) Sutra and the Lopana Sthapana Sutra and by merely looking at the frightful looking (but really harmless) Quadratic before us, readily by mere mental arithmetic, write down the answer to this question and sa y :— “ Yes ; and the straight lines are 3 x —2 y + 7 = 0 and 4 x + 5 y — -5=0. How exactly we do this (by mental arithmetic), we proceed to explain presently. The Vedic Method ( 1 ) By the “ Urdhva Tiryak” , the 3 x —2 y + 7 “ Lopana Sthapana” and th%“ Adyam Adyena” 4 x + 5 y —5 Sutras (as explained in some of the 12x2 + 7 x y —lOy v . V x \ „ + 1 3 x + 4 5 y -3 5 earliest chapters), we have (mentally): -------------- ---------12x2 —7xy+10y 2 = (3 x —2y) (4x+5y) and we find 7 and—5 to be the absolute of the two factors. We thus get (3x—2 y + 7 )= 0 and (4 x + 5 y —5 )= 0 as the two straight lines represented by the given equation. And that is all there is to it. The Hyperbolas and the Asymptotes. Dealing with the same principle and adopting the same procedure in connection with the Hyperbola, the Conjugate Hyperbola and the Asymptotes, in articles 324 and 325 on pages 293 & 294 of his “ Elements of Co-ordinate Geometry” , Prof. S. L. Loney devotes 27+14 (= 41) lines in all to the problem and concludes by saying: “ As 3x 2 —5 x y --2 y * + 5 x + lly —8 Asymptotes,

= 0

is the equation of the

/ . c = —12 The Equation to the Asymptotes is 5 x + l l y —1 2 = 0

3

x 2 —5 xy—2 y2+

And consequently the Equation to the Conjugate Hyper bola is 3x2—5xy—2y2+ 5 x + l l y —16. W ell; all this is not so terrific-looking, because of the very simple fact that all the working (according to Art, 116 on pages 9 5 etc.,) has been taken fpr granted and done “ out of Court”

( 369 ) or in private, so to speak. But even then the substitution of the values of a, b, c, f, g, and K in the Discriminant to the General Equation and so on is, from the Yedic standpoint, wholly supererogatory toil and therefore to be avoided. By the Vedic method, however, we use the same Lopan Sthdpana the Urdhva Tiryak and the Adyam Adyena Sutras; we first get-mentally 3 x + y and x —2 y and then—4 and 3 as the

3x-f-y — 4 x —2 y - } - 3

only possibilities in the case; and 3 x 2 _ 5 xy _ 2 y 2 + 5 x + l l y — 1 2 as this givesus- 1 2 in the product, --------------- --------------------------we get this product= 0 as the Equation to the Asymptotes; and, as the Conjugate Hyperbola is at the same distance-in the opposite direction from the Asymptotes, we put down the same equation (with only—16 instead of—8 ) as the required Equation to the Conjugate Hyperbola (and have not got to bother about the complexities of the Discriminants, the inevitable substi tutions and all the rest of i t ) ! And that is all. A few more illustrative instances will not be out of place: (1)

8

x 8 + ! 0 x y —3y 2 —2 x -f4 y —2 = 0 ,\ (2x-)-3y)(4x—y )—2 x + 4 y —2

= 0

1 J

4x— y + 1 2 x + 3 y —1______________ 8xa-f-10xy—3ya—2 x + 4 y —1 = 0 The Equation to the Asymptotes is 8 xa-f-1 0 x y —3ya—2 x -f-4 y = l ; and the Equation to the Conjugate Hyperbola is 8

x a+

1 0

x y —3y2 —2 x + 4 y = 0

(2 ) ya—x y —2 xa—5 y + x —6

= 0

y + x —2 " 1 y —2x—3 J _______________ y 2—x y —2 x 2 - f x —5 y + 6 = 0

The Asymptotes are ( y + x —2 ) (y—2x—3 )= 0 \ And the Conjugate Hyperbola is y 2—x y —2x2+ x —5 y + 1 8 = 0 J (3) 55x2—120xy+20y2+ 6 4 x —48y=0 l l x — 2y+4 5x—10y+4 5 5

x 2—120xy+20y8+ 6 4 x —48y+ 16 = 0

( 360 ) / . This is the Equation to the Asymptotes; and the Equation to the Conjugate Hyperbola is 55x2—120xy+20y2+ 6 4 x —48y+ 3 2= 0 (4) 12xs—23xy+10ya—25x+26y—14 ,\ The Asymptotes are :

*

4x—5y—3 3x—2y—4

(1 2 x 2—23xy+10y2—25x+26y

\

+ 12=0

%

And the Conjugate Hyperbola is )1 2 x 2—-23xy+10y2—25x+26y (. +38=0 (5) 6x2—oxy—6y2+ 1 4 x + 5 y + 4 .\ 2x—3 y + 4 ) 3 x -2 y + l { Independent term =4 Two straight lines.

)

Chapter XL. MISCELLANEOUS MATTERS There are also various subjects of a miscellaneous character which are of great practical interest not only to mathematicians and statisticians as such but also to ordinary people in the ordinary course of their various businesses etc., which the modern system of ing etc., .does scant justice to and in which the Vedic Sutras can be very helpful to them. We do not propose to deal with them now, except to name a few of them: ( 1 ) Subtractions; (2 ) Mixed additions and subtractions ; (3) Compound additions and subtractions* * (4) Additions of Vulgar Fractions e t c ; (5 ) Comparison of Fractions; ( 6 ) Simple and compound practice (without Aliquot parts etc.)

taking

(7) Decimal Operations in all Decimal W ork ; (8 ) Ratios, Proportions, Percentages, Averages etc. ; (9) Interest; Annuities, Discount e t c ; ( 1 0 ) The Centre of Gravity of Hemispheres etc ; ( 1 1 ) Transformation of Equations; and ( 1 2 ) Dynamics, Statics, Hydrostatics, Pneumatics etc., Applied Mechanics etc., etc. N.B. :—There are some other subjects, however, of an important character which need detailed attention but which (owing to their being more appropriate at a later stage) we do not now propose to deal with but which, at the same time, in view of their practical importance and their absorbingly interesting character, do require a brief description. We deal with them, therefore, briefly hereunder.

( 302 ) Solids, Trigonometry, Astronomy Etc. In Solid Geometry, Plane Trigonometry, Spherical Tri gonometry and Astronomy too, there are similarly huge masses of Vedic material calculated to lighten the mathematics students’ burden. We shall not, however, go here and now into a detailed disquisition on such matters but shall merely name a few of the important and most interesting headings under which these subjects may be usefully sorted : ( 1 ) The Trigonometrical Functions and their inter relationships ; etc. (2 ) Arcs and chords of circles, angles and sines of angles etc ; (3 ) The converse i.e. sines of angles, the angles themselves, chords and arcs of circles etc ; (4) Determinants and their use in the Theory of Equa tions, Trigonometry, Conics, Calculus etc ; ( 5 ) Solids and why there can be only five regular Poly hedrons; etc., etc. ( 6 ) The Earth’s daily Rotation on its own axis and her annual relation around the Sun; (7) Eclipses; ( 8 ) The Theorem (in Spherical Triangles) relating to the product of the sines of the Alternate Segments i.e. about : Sin BD Sin CE Sin AF__ , Sin DC ’ Sin EA Sin FB (9 ) The value of 1 1 (i.e. the ratio of the circumference of a circle to its Diameter). N.B. ; —The last item, however, is one which we would like to explain in slightly greater detail. Actually, the value of i i is given in the well-known Anustub metre and is couched in the Alphabetical Code-Language (described in an earlier chapter):

( 363 ) It is so worded that it can bear three different meanings— all o f them quite appropriate. The first is a hymn to the Lord Sri Krsna ; the second is similarly a hymn in praise of the Lord Shri Shankara: and the third is a valuation of — to 32 10 places of Decimals! (with a “ Self-contained master-key” for extending the evaluation to any number of decimal places ! As the student (and especially the non-Sanskrit knowing student) is not likely to be interested in and will find great difficulty in understanding the puns and other literary beauties of the verse in respect of the first two meanings but will naturally feel interested in and can easily follow the third meaning, we give only that third one here :

r-* ^ .3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3 ) 2 3 8 4 6 2 6 4 3 3 8 3 2 7 92. on which, on understanding it, Dr. V. P. Dalai (of the Heidelburg University, ) felt impelled—as a mathematician and physicist and also as a Sanskrit scholar—to put on record his comment as follows: 1 1 10

“ It shows how deeply the ancient Indian mathe maticians penetrated, in the subtlety of their calculations, even when the Greeks had no numerals above 1 0 0 0 and their multiplications were so very complex, which they performed with the help of the counting frame by adding so many times the multiplier ! 7 x 5 could be done by adding 7 on the counting frame 5 tim es!” etc., etc. !

RECAPITULATION AND CONCLUSION In these pages, we have covered a large number of branches of mathematics and sought, by comparison and contrast, to make the exact position clear to all seekers after knowledge. Arith metic and Algebra being the basis on which all mathematical operations have to depend, it was and is both appropriate and inevitable that, in an introductory and preliminary volume of this particular character, Arithmetic and Algebra should have received the greatest attention in this treatise. But this is only a kind of preliminary “PROLEGOMENA” and SAMPLE type of publication and has been intended solely for the purpose of giving our readers a foretaste of the delicious delicacies in store for them in the volumes ahead.1 If this volume achieves this purpose and stimulates the reader’s interest and prompts him to go in for a further detailed study of Vedic Mathematics we shall feel more than amply rewarded and gratified thereby.

1 No subsequent volume has been left by the author.—Editor.

A REPRESENTATIVE PRESS OPINION Reproducedfrom the Statesman, India, dated 10th January, 1956. EVERY MAN A MATHEMATICIAN (M b . D e s m o n d D o ig )

Now in Calcutta and peddling a miraculous commodity is His Holiness Jagad Guru Sri Shankaracharya of the Govardhan Peeth, Puri. Yet Sri Shankaracharya denies any spiritual or miraculous powers, giving the credit for his revolutionary knowledge to anonymous ancients who in 16 Sutras and 120 words laid down simple formulae for all the world’s mathematical problems. The staggering gist of Sri Shankaracharya’s peculiar knowledge is that he possesses the know-how to make a mathe matical vacuum like myself receptive to the high voltage of higher mathematics. And that within the short period of one year. To a person who struggled helplessly with simple equa tions ahd simpler problems, year after school-going year and without the bolstering comfort of a single credit in the subject, the claim that I can face M.A. Mathematics fearlessly after only six months of arithmetical acrobatics, makes me an im mediate devotee of His Holiness Jagad Guru Sri Shankaracharya of the Goverdhan Peeth, Puri. I was introduced to him in a small room in Hastings, a frail but young 75 year-old, wrapped in pale coral robes and wearing light spectacles. Behind him a bronze Buddha caught the rays of a tresing sun, splintering them into a form of aura; and had ‘Bis Holiness’ claimed divine inspiration* I would have belived him. He is that type of peison, dedicated as, much as I hate using the word ; a sort of saint in saint’s clothing, and no inkling of anything so mundane as a mathe matical mind.

( 366 )

Astounding Wonders. My host, Mr. Sitaram, with whom His Holiness Sri Shankaracharya is staying, had briefly* prepared me for the interview. I could pose any question I wished, I could take photographs, I could read a short descriptive note he had prepared on “ The Astounding Wonders of Ancient Indian Vedic Mathematics” . His Holiness, it appears, had 'Spent years in contemplation, and while going through the Vedas had suddenly happened upon the key to what many historians, devotees and translators had dismissed as meaningless jargon. There, contained in certain Sutras, were the processes of mathe matics, psychology, ethics, and metaphysics. “ During the reign of King Kamsa” read a Sutra, “rebellions, arson, famines and insanitary conditions prevailed” . Decoded, this little piece of libellous history gave decimal answer to the fraction 1/17 ; sixteen processes of simple mathematics reduced to one. The discovery of one key led to another, ancj His Holiness found himself turning more and more to the astotmding know ledge contained in words whose real meaning had been lost to humanity for generations. This loss is obviously one of the greatest mankind has suffered; and, I suspect, resulted from the secret being entrusted to people like myself, to whom a square root is one of life’s perpetual mysteries. Had it survived, every-edumted-“ soul” ; would, be ; a, mathematical-‘ ‘wizard” ; and, maths-‘masters” would “ starve” . For my note reads “ Little children merely look at the sums written on the black board and immediately shout out the answers......they....... have merely to go on reeling off the digits, one after another forwards or backwards, by mere mental arithmetic (without needing pen, pencil, paper or slate).” This is the sort of thing one usually refuses to believe. I did. Until I actually met His Holiness. On a child’s blackboard, attended with devotion by .my host’s wife; His Holiness began demonstrating his peculiar

( 367 ) skill; multiplication, division, fractions, algebra, and intricate excursions into higher mathematics for which I cannot find a name, all were reduced to a disarming simplicity. Yes, I even shouted out an answer. (‘Algebra for High Schools5, Page 363, exercise 70, example ten). More, I was soon tossing off answers to problems, which ; official-Maths-books ; “ described” , a s ; “ advanced” , difficult, and very difficult. Cross my heart! His Holiness’s ambition is to restore this lost art to the world, certainly to India. That India should today be credited with having given the world, via Arabia, the present numerals we use, especially the epochmaking “ zero” , is not enough. India apparently once had the knowledge which we are today rediscovering. Somewhere along the forgotten road of history, calamity, or deliberate destruction, lost to man the secrets he had emassed. It might happen again. In the meantime, people like His Holiness Jagad Guru Sri Shankaracharya of the Govardhan Peeth, Puri, are by a devotion to true knowledge, endeavouring to restore to humanity an interest in great wisdom by making that wisdom more easily acceptable. Opposition there is, and will be. But eminent mathematicians both here and abroad are taking more than a ing interest in this gentle ascetic’s discoveries. I for one, as a representative of all the mathematically despairing, hope, sincerely hope, that his gentle persuasion will prevail.

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