# 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 $abc|(a+b+c)^{31}$.

5 years, 1 month 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, 1 month ago

Here a , b, c are becoming equal no?

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month 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, 1 month 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, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

This seems fairly easy for an Olympiad problem

- 4 years, 7 months ago