पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३५०

154

GAŅITASĀRASAŃGRAHA.

multiplying these sums with each other, giving up in each case the sum relating to the particular specified multiple, are to be reduced to their lowest terms by the removal of common factors. (These reduced quantities are then) to be added. (Thereafter) the square root (of this resulting sum) is to be obtained, from which one is (to be subsequently) subtracted. Then the reduced quantities referred to above are to be multiplied by (this) square root as diminished by one. Then these are to be separately subtracted from the sum of those same reduced quantities. Thus the moneys on hand with each (of the several persons) are arrived at . These (quantities measuring the moneys on hand) have to be added to one another, excluding from the addition in each case the value of the money on the hand of one of the persons; and the several sums so obtained are to be written down separately. These are (then

 

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Then, ${\tfrac {(n+1)(b+1)(c+1)}{T}}\times (y+z)=(b+1)(c+1)$ . . . . .	I
where $T=u+x+y+z.$
Similarly, ${\tfrac {(a+1)(b+1)(c+1)}{T}}\times (z+x)=(c+1)(a+1)$ . . . . .	II
and         ${\tfrac {(a+1)(b+1)(c+1)}{T}}\times (x+y)=(a+1)(b+1)$ . . . . .	III
Adding, I, II, III,
${\begin{aligned}{\tfrac {(a+1)(b+1)(c+1)}{T}}&\times 2(x+y+z)=(b+1)(c+1)\\&+(c+1)(a+1)+(a+1)(b+1)\\&=S(say)\end{aligned}}$ . . . . .	IV

Subtracting separately I, II, III,

${\begin{aligned}&{\tfrac {(a+1)(b+1)(c+1)}{T}}\times 2x=S-2(b+1)(c+1).\\&{\tfrac {(a+1)(b+1)(c+1)}{T}}\times 2y=S-2(c+1)(a+1)\\&{\tfrac {(a+1)(b+1)(c+1)}{T}}\times 2x=S-2(a+1)(b+1)\end{aligned}}$

$\therefore x:y:z::S-2(b+1)(c+1):S-2(c+1)(a+1):S-2(a+1)(b+1).$

By removing the common factors, if any, in the right-hand side of the proportion, we get at the smallest integral values of x,y,z.

This proportion is given in the rule as the formula.

It may be noted that the square root mentioned in the rule has reference only to the problem given in the stanzas 236-287. Correctly speaking, instead of "square root", we must have '3'.

It can be seen easily that this problem is possible only when the sum of any two of ${\tfrac {1}{a+1}},{\tfrac {1}{b+1}},{\tfrac {1}{c+1}}$ greater than the third.