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December 3, 2012
RMO 2012 Solution to Question No. 2
2. Let a, b, c be positive integers such that a divides , b divides and c divides . Prove that abc divides . Solution: A general term of the expansion of is 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 : We will show that is divisible by c where p+q=31 Suppose then c divides as it contains and by problem c divides Again if p-1 < 5 then as p-1 + q-1 = 29 (as p+q = 31) Now a divides or divides . As c divides and divides hence c divides implying c divides as 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. . c divides and b divides which divides . Hence we have checked all possible terms and have shown than abc divides each of them.