# Indian Regional Mathematical Olympiad 2013(problem 2)

If $a$,$b$ and $c$ are positive integers such that $a | b^{5}$ , $b|c^{5}$ and $c|a^{5}$.Prove that $abc|(a+b+c)^{31}$.

Note by Eddie The Head
5 years, 6 months ago

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in general if $a | b^n$ , $b | c^n$ , $c | a^n$ , $abc | (a+b+c)^{n^2 + n + 1}$

- 5 years, 6 months ago

Here a , b, c are becoming equal no?

- 5 years, 6 months ago

$a,b,c$ are not necessarily equal, if that's what you mean.

- 5 years, 6 months ago

You re correct ...at first glance it looked like that....:p I sometimes suffer from brain malfunctions like these.....

- 5 years, 6 months ago

This was my favorite question of the set. It looks so elegant, and is simple to approach if you are careful with the details.

Staff - 5 years, 6 months ago

Outline:

Consider each term of the expansion of $(a+b+c)^{31},$ which is of the form $m a^{p} b^{q} c^{r},$ where $m, a, b, r \in \mathbb{Z^+}$ and $p+q+r=31.$ If $p,q,r>0,$ $abc \mid a^p b^q c^r.$ You just need to handle the exceptional cases when atleast one of $p,q,r$ is zero, which is not hard. I'll post my full solution if necessary.

I got this problem when I sat for RMO last time. Heaven knows why I couldn't solve it in the examination hall... it was pretty easy.

- 5 years, 6 months ago

I solved this problem orally with you while coming down the staircase :P

- 5 years, 6 months ago

Yep, I remember! :P I couldn't solve it in the hall, though.

- 5 years, 6 months ago

This seems fairly easy for an Olympiad problem

- 5 years ago