# पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३१५

एतत् पृष्ठम् परिष्कृतम् अस्ति
119
CHAPTER VI--MIXED PROBLEMS.

given) divisor. (The remainder in this last division becomes the multiplier with which the originally given group-number is to be multiplied for the purpose of arriving at the quantity which is to be divided or distributed in the manner indicated in the problem. Where, however, the given group-numbers, increased or decreased in more than one way, are to be divided or distributed in more than one proportion,) the divisor related to the larger group-value,(arrived at as explained above in relation to either of two specified distributions), is to be divided (as above) by the divisor (related to

By choosing a value for p4, such that ${\displaystyle {\tfrac {r_{5}p_{4}+b}{r_{4}}}}$, which is, as shown above, the value of p4,becomes an integer, and by arranging in a chain ${\displaystyle q_{2},q_{3},q_{4},q_{5},p_{4}{\text{ and }}p_{5}}$, we get at the value of x by proceeding as stated in the rule, that is, by the processes of multiplication by the upper quantity and the addition of the lower quantity in the chain, which are carried up to the topmost quantity. The value of x so obtained is divided by A, and the remainder represents the least value of x; for the values of x which satisfy the equation, ${\displaystyle {\tfrac {Bx+b}{A}}}$ an integer, are all in an arithmetical progression wherein the common difference is A.

This same rule contemplates problems where two or more conditions are given, such as the problems given in stanzas 121${\displaystyle {\tfrac {1}{2}}}$ to 129${\displaystyle {\tfrac {1}{2}}}$. he problem in 121${\displaystyle {\tfrac {1}{2}}}$ may be thus worked out according to the rule:- It is given that a heap of fruits when diminished by 7 is exactly divisible among 8 men, and the same heap when diminished by 3 is exactly divisible among 13 men.

Now, according to the method already given, find out first the least number of fruits that will satisfy the first condition, and then find out the number of fruits that will satisfy the second condition. Thus we get the group-values 15 and 16 respectively. Now, the divisor related to the larger group-value is divided as before by that related to the smaller group-value to obtain a fresh vallikā chain. , Thus dividing 13 by 8 and continuing the division, we have:-

 8)13(1      8     ----      5)8(1         5        ----         3)5(1            3           ----            2)3(1               2              ----               1)2(1                  1                  --                  1 1 1 1 1 1 2

From this the Vallikā chain comes out thus:- Choosing 1 as the mati, and adding the difference between the two group-values already arrived at, that is, 16–15, or 1, to the product of the mati and the last divisor, and dividing this sum by the last divisor, we have 2, which is to be written down below the mati in the Vallikā chain. Then proceeding as before with the vallikā, we get 11, which, when divided by the first divisor 8, leaves the remainder 3. This is multiplied by the divisor related to the larger group-value, viz., 13, and then is combined with the larger group-value. Thus 65 is the number of fruits in the heap.