RMO 2012 Solution to Question No. 2

2. Let a, b, c be positive integers such that a divides \((b^5)\) , b divides \((c^5)\) and c divides \((a^5)\). Prove that abc divides \(((a+b+c)^{31})\).


A general term of the expansion of \(((a+b+c)^{31})\) is \((\frac {31!}{p!q!r!} a^p b^q c^r)\) where p+q+r = 31 (by multinomial theorem; this may reasoned as following: from 31 factors (a+b+c), choose p factors and from those chosen p factors take out ‘a’. From remaining 31-p factors choose q factors and from these chosen q factors take out ‘b’. From the remaining r factors take out ‘c’.)

Now the terms of the expansion can be of three types:

Case 1

p, q and r are all non-zero. These terms are straight away divisible by abc as all of a, b, and c are present in them.

Case 2

Exactly one of p, q, r is zero and rest two are non-zero. Let us examine the subcase where r=0 and p,q are non zero. Other two subcases will be similar.

Then p+q+0 = 31 or p+q=31

Term of the expansion will have :

\((a^p b^q = ab(a^{p-1} b^{q-1}))\)

We will show that \((a^{p-1} b^{q-1})\) is divisible by c where p+q=31

Suppose \(((p-1)\ge 5 , (q-1)\ge 5)\) then c divides \((a^{p-1})\) as it contains \((a^5)\)  and by problem c divides \((a^5)\)

Again if p-1 < 5 then \(((q-1) \ge 25 )\) as p-1 + q-1 = 29 (as p+q = 31)

Now a divides \((b^5 )\) or \((a^5)\) divides \((b^{25})\). As c divides \((a^5)\) and \((a^5)\) divides \((b^{25})\) hence c divides \((b^{25})\) implying c divides \((b^{q-1})\) as \(((q-1)\ge 25 )\)

Case 3.

Exactly two of p, q and r are zero. Let us again examine of the three subcases where q=0, r=0 and p nonzero. Other two subcases will be similar.

Then p = 31.

\((a^{31} = a\times a^5 \times a^{25})\). c divides \((a^5)\) and b divides \((c^5)\) which divides \((a^{25})\).

Hence we have checked all possible terms and have shown than abc divides each of them.

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