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# 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 $((a+b+c)^{31})$.

Solution:

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.