Indian Regional Mathematical Olympiad 2013(problem 2)

If aa,bb and cc are positive integers such that ab5a | b^{5} , bc5b|c^{5} and ca5c|a^{5}.Prove that abc(a+b+c)31abc|(a+b+c)^{31}.

#NumberTheory #CosinesGroup #Goldbach'sConjurersGroup #TorqueGroup

Note by Eddie The Head
7 years, 2 months ago

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Comments

in general if abna | b^n , bcnb | c^n , canc | a^n , abc(a+b+c)n2+n+1abc | (a+b+c)^{n^2 + n + 1}

Sagnik Saha - 7 years, 2 months ago

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Here a , b, c are becoming equal no?

Eddie The Head - 7 years, 2 months ago

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

Daniel Liu - 7 years, 2 months ago

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

Eddie The Head - 7 years, 2 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.

Calvin Lin Staff - 7 years, 2 months ago

Outline:

Consider each term of the expansion of (a+b+c)31,(a+b+c)^{31}, which is of the form mapbqcr,m a^{p} b^{q} c^{r}, where m,a,b,rZ+m, a, b, r \in \mathbb{Z^+} and p+q+r=31.p+q+r=31. If p,q,r>0,p,q,r>0, abcapbqcr.abc \mid a^p b^q c^r. You just need to handle the exceptional cases when atleast one of p,q,rp,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 - 7 years, 2 months ago

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

Sagnik Saha - 7 years, 2 months ago

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

Sreejato Bhattacharya - 7 years, 2 months ago

This seems fairly easy for an Olympiad problem

Aritra Jana - 6 years, 8 months ago
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