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}.

Note by Eddie The Head
5 years, 6 months ago

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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 - 5 years, 6 months ago

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Here a , b, c are becoming equal no?

Eddie The Head - 5 years, 6 months ago

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

Daniel Liu - 5 years, 6 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 - 5 years, 6 months ago

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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.

Calvin Lin Staff - 5 years, 6 months ago

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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 - 5 years, 6 months ago

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

Sagnik Saha - 5 years, 6 months ago

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

Sreejato Bhattacharya - 5 years, 6 months ago

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This seems fairly easy for an Olympiad problem

Aritra Jana - 5 years ago

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