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.

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## Comments

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

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Here a , b, c are becoming equal no?

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

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

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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.

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

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

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This seems fairly easy for an Olympiad problem

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