119 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 on 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 Pa such that shown above, the value of ps, becomes an integer, and by arranging in a cham 42, 43, 44, 45, P4 and ps we get at the value of a 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 so obtained is divided by A, and the remainder represents the least Be + b value of a, for the values of a which satisfy the equation, =an integer, A are all in an arithmetical progression wherein the common difference is 4. This same rule contemplates problems where two or more conditions are given, such as the problems given in stanzas 121 to 129, The problem in 1213 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 henp when diminished by 3 is exactly divisible among 13 mon. 5)8(1 ŠŠŠŠ Now, according to the method already given, find out first the least number of fruits that will satisfy the fist 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 fiesh rallka chain. Thus dividing 13 by 8 and continuing the division, we have- 8)13(1 8 3)5(1 CHAPTER VI-MIXED PROBLEMS. 2) 3 ( 1 1)2(1 1 1 14 P₁+ b 14 From this the Valhká chain comes out tlus - Choos- ing 1 as the math, and adding the difference between the two group-values already arrived at, that is, 16-15, or 1, to the product of the mats and the last diviso1, and dividing this sum by the last divisor, we have 2, which is to be written down below the mati in the Valliká chain. Then pro- ceeding as before with the vallakä, 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 55 is the number of fruits in the heap. 1 1 1 , which is, 2
