THEORY OF THE PULVERISER 109 गुणकारगुणे शेषे लब्धगुणे हारभाज्यसंहृतयोः । शेषौ तत्र क्रमशो दिनचक्रादी भवेतां तौ ॥ १३ ॥1 etc 2.८., “or (alternatively), solve the pulveriser by taking +! (if the given interpolator is positive) or - 1 (if the given inter p0lator is negative). The remainders (resulting in this way) from the upper and lower numbers (of the reduced chain) are the (corresponding) multiplier and quotient (respectively) Multiply the (given) interpolator severally by these multiplier and quotient and divide (the products thus obtained) by the divisor and the dividend (respectively). The remainders (thus obtained) are respectively the days, etc., and the revolutions To get a solution of Ex. 3 by this method, we solve the pulveriser 36641 x - ॥ 394479375 by the previous method, and get 109 ४=113065211 (multiplier), y=10502 (पuotient) Now multiplying 113065211 by 2+ and dividing the product by 394479375, we get 346688814 as remainder. These are the reguired days. Again multiplying 10502 by 24 and dividing the product by 36641, we get 32202 as remainder. These are the required revolutions of Saturn. This method is based on the consideration that if*=4, y८ B be a solu tion of (८४:+1)| b=y, then ४=64, 1y =cB will be a solution of (८४+८)|b= y The importance of this method lies in the fact that any astronomical problem like the one considered above may be solved by talking recourse to the table of solutions of the euations (0.४:+1)/b =y, for different values of a and b.
