पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३१४

118

GAŅITASĀRASAŃGRAHA.

with which the least remainder in the odd position of order (in the above-mentioned process of successive division ) is to be multiplied; and (then put down) below (this again) this product increased or decreased (as the case may be by the given known number) and then divided (by the last divisor in the above mentioned process of successive division. Thus the Vallikā or the creeper-like chain of figures is obtained. In this) the sum obtained by adding (the lowermost number in the chain) to the product obtained by multiplying the number above it with the number (immediately) above (this upper number, this process of addition being in the same way continued till the whole chain is exhausted, this sum, is to be divided by the (originally

 

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1--51
2--38
1--13
4--12
1
8

Thus we get the chain or Vallikā noted in the first column of figures in the margin. Then we multiply the penultimate figure below in the chain, viz., 1, by 4, which is above it, and add 8, the last number in the chain; the resulting 12 is written down so as to be in the place corresponding to 4; then multiplying this 12 by 1 which is the figure above it in the creeper chain, and adding 1, the figure similarly below it, we get 13 in the place of 1; proceeding in the same manner 38 and 51 are obtained in the places of 2 and 1 respectively. This 51 is divided by 28, the divisor in the problem ; and the remainder 5 is seen to be the length number of fruits in a bunch.

The rationale of the rule will be clear from the following algebraical representation:-

${\tfrac {Bx+b}{A}}=y{\text{ (an integer) }}=q_{1}x+p_{1},{\text{ where ) }}p_{1}={\tfrac {(B-Aq_{1})x+b}{A}}$.

$\therefore x={\tfrac {Ap_{1}-b}{r_{1}}}{\text{ , (where }}r_{1}=B-Aq_{1}{\text{ the first remainder) }}$

$=q_{2}p_{1}+p_{2}{\tfrac {}{}}$, where $p_{2}={\tfrac {r_{2}p_{1}-b}{r_{1}}}{\text{ and }}q_{2}{\text{ is the second quotient and }}r_{2}{\text{ the second remainder }}$

${\text{Hence,}}p1={\tfrac {r_{1}p_{2}+b}{r_{2}}}=q_{3}p_{2}+p_{3}{\text{, where }}p_{3}={\tfrac {r_{3}p_{2}+b}{r_{3}}}{\text{ and }}q_{3}{\text{ is the third quotient and }}r_{3}{\text{ the third remainder. }}$

${\text{Similarly, }}p_{2}={\tfrac {r_{2}p_{2}-b}{r_{2}}}=q_{4}p_{3}+p_{4}{\text{, where }}p_{4}={\tfrac {r_{4}p_{3}-b}{r_{2}}};$

$p_{2}={\tfrac {r_{2}p_{4}+b}{r_{4}}}=q_{5}p_{4}+p_{5}{\text{, where }}p_{5}={\tfrac {r_{5}p_{4}+b}{r_{4}}}$

Thus we have, $x=q_{2}p_{1}+p_{2};$

$p_{1}=q_{3}p_{2}+p_{2};$

$p_{2}=q_{4}p_{3}+p_{4};$

$p_{3}=q_{5}p_{4}+p_{5}.$