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Consider each term of the expansion of (a+b+c)31, which is of the form mapbqcr, where m,a,b,r∈Z+ and p+q+r=31. If p,q,r>0,abc∣apbqcr. 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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Comments
in general if a∣bn , b∣cn , c∣an , abc∣(a+b+c)n2+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.
Outline:
Consider each term of the expansion of (a+b+c)31, which is of the form mapbqcr, where m,a,b,r∈Z+ and p+q+r=31. If p,q,r>0, abc∣apbqcr. 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.
This seems fairly easy for an Olympiad problem