120 GANITASĀRASANGRAHA. the smaller group-value, so that a creeper-like chain of successive quotients may be obtained in this case also as before. Below the lowermost quotient in this chain, the optionally chosen multiplier of the least remainder in the odd position of order in this last successive division is to be put down; and below this again is to be put down the namber which is obtained by) adding the difference between the two group-values (already referred to) to the product (of the least remainder in the last odd position of order multiplied by the above optionally chosen multiplier thereof, and then by dividing the resulting sumr by the last divisor in the The rationale of this process will be clear from the following considerations - B₂ + b₂ is an integer, (n) 18 an mieger, and A₂ We have (1) B1 + b₁ 4₁ In (1) Let the lowest valne of = 81. x=80. In (1) In (m)
- =eg.
(iv) When both (1) and (11) are to be satisfied, dA, + 8, has to be equal to A₁ d + ($182) 1. KA₂ +82, so that s₁ s₂kA₂ - dA₁. That is, A₂ From (iv), which is an indeterminate equation with the values of d and k unknown, we arrive, according to what has been already proved, at the lowest positive integral value of d. This value of d multiplied by 4₁, and then increased by 8₁, gives the value of which will satisfy (1) and (n). Let this be ti, and let the next higher value of a which will satisfy both these equations be to. (v) Now, ti+nd₁ = t; (vi) and ti + mAg = t. 4₁ m ..As Thus 4₁ mp, and A, np, where p is the highest common factor between A₁ and 42. n B3 + b3 43 (iii) ..m= is an integer. -4, t₁ + .. , and n P Substituting in (v) or (vi), we have P From this it is obvious that the next higher value of satisfying the two equations is obtained by adding the least common multiple of 4, and A, to the lower value. Now again, let v be the valne of which satisfies all the three equations. A₁ A₂ xr, (where r 14 a positive integer) = (say) t₁ + Wr; P and v 83 + c3= t₁ + r. Then t₁ + CAST 3 + 5g - b 31 4₁ 4₂
