Tham khảo tài liệu 'đề thi toán apmo (châu á thái bình dương)_đề 10', khoa học tự nhiên, toán học phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | XIX Asian Pacific Mathematics Olympiad p. Problem 1. Let S be a set of 9 distinct integers all of whose prime factors are at most 3. Prove that S contains 3 distinct integers such that their product is a perfect cube. Solution. Without loss of generality we may assume that S contains only positive integers. Let S 2ai 3bi I a bi 2 Z ai bi 0 1 i 9 . It suffices to show that there are 1 i1 i2 i3 9 such that ai1 a 2 a 3 bii bi2 bi3 0 mod 3 . t For n 2a3b 2 S let s call a mod 3 b mod 3 the type of n. Then there are 9 possible types 0 0 0 1 0 2 1 0 1 1 1 2 2 0 2 1 2 2 . Let N i j be the number of integers in S of type i j . We obtain 3 distinct integers whose product is a perfect cube when 1 N i j 3 for some i j or 2 N i 0 N i 1 N i 2 0 for some i 0 1 2 or 3 N 0 j N 1 j N 2 j 0 for some j 0 1 2 or 4 N i1 j1 N i2 j2 N i3 j3 0 where i1 i2 i3 0 1 j2 j3 0 1 2 . Assume that none of the conditions 1 3 holds. Since N i j 2 for all i j there are at least five N i j s that are nonzero. Furthermore among those nonzero N i j s no three have the same i nor the same j. Using these facts one may easily conclude that the condition 4 should hold. For example if one places each nonzero N i j in the i j -th box of a regular 3 X 3 array of boxes whose rows and columns are indexed by 0 1 and 2 then one can always find three boxes occupied by at least one nonzero N i j whose rows and columns are all distinct. This implies 4 . Second solution. Up to t we do the same as above and get 9 possible types a mod 3 b mod 3 0 0 0 1 0 2 1 0 1 1 1 2 2 0 2 1 2 2 for n 2a3b 2 S. Note that i among any 5 integers there exist 3 whose sum is 0 mod 3 and that ii if i j k 2 0 1 2 then i j k 0 mod 3 if and only if i j k or i j kg 0 1 2 . Let s define T the set of types of the integers in S N i the number of integers in S of the type i M i the number of integers j 2 0 1 2 such that i j 2 T. If N i 5 for some i the result follows from i . Otherwise for some permutation i j k of 0 1 2 N i 3 N j 3 N k 1. If M i or M
