2014/051) Find smallest positive a such that 5x^13+13x^5 + 9ax is divisible by 65 for all x

let f(x) = x(5x^12 + 13x^4 + 9a)

x can be factored out so 65 should divide (5x^12 + 13x^4 + 9a)


So taking mod 5 we have 13x^4 + 9a   mod  5 = 0

It is true for any x so take x co-prime to 5 so we have x^4 = 1

So we get 13 + 9a  mod 5 = 0

or 3 – a mod 5 = 0

or a = 3 mod 5

Similarly we have (5x^12 + 13x^4 + 9a) mod 13 = 0

 or 5 + 9a mod 13 = 0

or a = 11 mod 13

a = 3 mod 5 and a = 11 mod 13 can be solved by using Chinese  remainder theorem also but we see that
a = -2 mod 5 and a = -2 mod 13 so a = -2 mod 65 or 63

a = 63 + 65n and a = 63 is the lowest positive integer

you can find some discussion at http://mathhelpboards.com/challenge-questions-puzzles-28/divisibility-challenge-10574.html#post49059