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

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TopNewestin general if \(a | b^n\) , \(b | c^n\) , \(c | a^n\) , \(abc | (a+b+c)^{n^2 + n + 1}\) – Sagnik Saha · 2 years, 11 months ago

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– Eddie The Head · 2 years, 11 months ago

Here a , b, c are becoming equal no?Log in to reply

– Daniel Liu · 2 years, 11 months ago

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

– Eddie The Head · 2 years, 11 months ago

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

This was my favorite question of the set. It looks so elegant, and is simple to approach if you are careful with the details. – Calvin Lin Staff · 2 years, 11 months ago

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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. – Sreejato Bhattacharya · 2 years, 11 months ago

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– Sagnik Saha · 2 years, 11 months ago

I solved this problem orally with you while coming down the staircase :PLog in to reply

– Sreejato Bhattacharya · 2 years, 11 months ago

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

This seems fairly easy for an Olympiad problem – Aritra Jana · 2 years, 5 months ago

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