A = (225a^2 + 45a + 10) / a; Where A belongs to integar. How many integral solutions of A are possible?

Question

A = (126a ^ 2 + 113a + 11)/a;
Where A belongs to integer.
How many integral solutions of A are possible?

Answer 1

Four

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In order A gets integer value all the terms in the parenthesis must be divisible by a.
The first 2 terms are divisible by a as have a as a factor
The third term, 10 must be divisible by a.
Since 10 has prime factors of 1, 2, 5, 10 there are four solutions only.