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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.

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