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Given 33 natural number so that their prime divisor just with $ 7,5,2,3,11$ is formed. Prove that multiplication two number of these numbers are complete square.

Thank you.

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    @gt6989b: What product? I don't understand.2012-12-13

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Hint: Recall that the positive integer $n$ is a perfect square if and only if in the prime power factorization of $n$, each exponent is even.

Any number whose prime divisors do not include any primes other than the ones mentioned can be written as $k^2(2^a3^b5^c7^d11^e)$ where $a, b,c,d,e$ are $0$ or $1$. There are only $2^5$ sequences of $0$'s and/or $1$'s of length $5$. Now use the Pigeonhole Principle.

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    Well, it is immediate that for each of our $5$ primes, the only thing that matters is the parity of the exponent.2014-11-17
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Hint: you need the product to have an even number of powers of each of the primes. How many combinations of odd/even powers are possible?

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    @misi10: it is like flipping a coin. If you had only one prime, there would be 2 possibilities: odd or even. If you had two primes, there would be 4: EE, OE, EO, OO. Can you keep going?2012-12-13