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Prove that for each 201 number that we will choose from [1,300] there will be always some x,y that thiers division will be power of 3.

For example:

$18 / 6$

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1 Answer 1

Suppose you had $201$ distinct integers in $[1,300]$ such that none is a power of $3$ times another. If the least of these $\le 100$, multiply it by $3$ and the property is still satisfied. Repeat until there are no more numbers $\le 100$. But there are only $200$ integers from $101$ to $300$.

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We can assume that your answer is right,but yesterday I had an exam and that was one of the question that I had to answer on and that will be lovely if you will post a full answer. –  Gil Sep 30 '13 at 15:07
    
Hopefully we can assume that your instructor reads math.stackexchange.com. –  Robert Israel Sep 30 '13 at 15:11
    
But it would be even lovelier if you would tell me his or her email address, so I could send it directly and save you the bother of copying the answer yourself. –  Robert Israel Sep 30 '13 at 15:32

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