पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३२३

127

CHAPTER V--MIXED PROBLEMS.

numbers immediately above it, (till the topmost figure in the chain becomes included in the operation), is to be arrived at. (Thereafter) this resulting sum and the divisor in the problem (give rise), in the shape of two remainders, (to the two values of) the unknown quantity (which is to be multiplied by the given dividend-coefficient in the problem),which (values)are related either to the known given quantity that is to be added or to the known given quantity that is to be subtracted, according as the number of figure-links in the above-mentioned chain of quotients is even or odd. (Where, however, the given groups, 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 over and over (as above by the divisor

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Carry out, the required process of continued division:--

67)59(0        0       ----       59)67(1            59           ----             8)59(7                56               -----                  3)8(2                     6                   -----                     2)3(1                        2                       ----                        1)2(1                           1                          ----                           1

After discarding the first quotient, the others are written down in a chain thus:-- 1 7 2 1 1 1 1+13=14 Below this are next written down 1 and 1, the last equal divisor and remainder. Here also, as in Vallikā-kuṭṭīkāra is worthy of note that in the last division there can be really no remainder, as 2 is fully divisible by 1 . But since the last remainder is wanted for the chain, it is allowed to occur by making the last quotient smaller than possible. And to the last number 1 here, add 13, which is the remainder obtained by dividing 80 by 67; the 14 so obtained is also written down at the bottom of the chain,which now becomes complete.

1-392
7-345
2-47
1-16
1-15
1
14

Now, by the continued process of multiplying and adding the figures in this chain, as already explained in the note under stanza No. 115${\tfrac {1}{2}}$. This is then divided by 67; and the remainder 57 is one of the values of x, when 80 is taken as negative owing to the number of figures in the chain being odd. When 80 is taken as positive, the value of x is 67 - 57 or 10. If the number of figures in the chain happen to be even, then the value of x first arrived at is in relation to the positive agra; if this value be subtracted from the divisor, the value of x in relation to a negative agra is arrived at.