AoPSWiki
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
Personal tools

1985 IMO Problems/Problem 4

From AoPSWiki

Problem

Given a set M of 1985 distinct positive integers, none of which has a prime divisor greater than 23, prove that M contains a subset of 4 elements whose product is the 4th power of an integer.

Solution

We have that x\in M\Rightarrow x=2^{e_1}3^{e_2}\cdots 19^{e_8}23^{e_9}. We need only consider the exponents. First, we consider the number of subsets of two elements, such that their product is a perfect square. There are 2^9=512 different parity cases for the exponents e_1,e_2,...,e_9. Thus, we have at least one pair of elements out of 1985 elements. Removing these two elements yields 1983 elements. By applying the Pigeon Hole Principle again, we find that there exists another such subset. Continuing on like this yields at least 734 pairs of elements of M whose product is a perfect square. Let the S be the set of the square root of the product of each pair. Then, by the Pigeon Hole Principle again, there exist at least two elements whose product is a perfect square. Let the elements be x,y and let x=\sqrt{ab},y=\sqrt{cd} where a,b,c,d\in M. Then, we have xy=z^2 for some z which implies abcd=z^4 and the claim is proved.

See also

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us