पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३२०

124

GAŅITASĀRASAŃGRAHA.

the specified known quantity which is to be taken away (from the previous remainder) is added: (and this resulting sum) is multiplied by that (same kind of) remaining fractional proportion (of the remainder as has been mentioned above). This is to be done as many times as there are distributions to be made. Then these quantities so obtained should be deprived of their denominators; and these denominator-less quantities (and the successive products of the above-mentioned remaining fractional proportions of the remainder) are (to be used as) the known quantity and the (other elements, viz., the coefficient) multiple (of the unknown quantity and the divisor, required in relation to a problem on Vallikā-kuṭṭikāra).

Examples in illustration thereof.

131${\tfrac {1}{2}}$. On a certain man bringing mango fruits (home, his) elder son took one fruit first and then half of what remained (On the elder son going away after doing this), the younger (son) did similarly (with what was left there. He further took half

 

---

 

The rule will be clear from the following working of the problem in 132${\tfrac {1}{2}}$-133${\tfrac {1}{2}}$:-

Here 1 is the first agra, and ${\tfrac {1}{3}}$ is the first agrāṁśa; therofore $1-{\tfrac {1}{3}}$ or ${\tfrac {2}{3}}$ is the śēșāṁśa. Now, obtain the product of agra and śēșāṁśa or $1\times {\tfrac {2}{3}}$ or ${\tfrac {2}{3}}$. Write it down in 2 places. ${\begin{cases}{\tfrac {2}{3}}\\{\tfrac {2}{3}}\end{cases}}{\Bigg \}}$                

I

Repeat the quantities ${\begin{cases}{\tfrac {2}{3}}\\{\tfrac {2}{3}}\end{cases}}{\Bigg \}}$; add the second agra 1(to one of the quantities) Then we have${\begin{cases}{\tfrac {5}{3}}\\{\tfrac {2}{3}}\end{cases}}{\Bigg \}}$;

multiply both by the next śēșāṁśa $1-{\tfrac {1}{3}}$ or ${\tfrac {2}{3}}$, so that you get ${\begin{cases}{\tfrac {10}{9}}\\{\tfrac {4}{9}}\end{cases}}{\Bigg \}}$               

II

Take these figures and add the third agra 1 as before; and you have ${\begin{cases}{\tfrac {10}{9}}\\{\tfrac {4}{9}}\end{cases}}{\Bigg \}}$; multiply by the next śēșāṁśa $1-{\tfrac {1}{3}}$ or ${\tfrac {2}{3}}$ and by the last aṁsa or ${\tfrac {1}{3}}$ and you have ${\begin{cases}{\tfrac {36}{81}}\\{\tfrac {8}{81}}\end{cases}}{\Bigg \}}$

III

The denominators of the first fractions in these three sets of fractions marked I, II, III, are dropped, and the numerators represent negative agra in a problem on Vallikā-kuṭṭīkāra, wherein the numerator and the denominator of each of the second fractions in those sets represent respectively the dividend coefficients and the divisor. Thus we have

${\tfrac {2x-2}{3}}$ is an integer; ${\tfrac {4x-10}{9}}$is an integer; and ${\tfrac {8x-38}{81}}$ is an integer.

The value of a satisfying these three conditions gives the number of flowers.