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Can anyone help me prove what is the smallest $n \in \mathbb{N}$ such that $n$ is divisible by $2,3,5$, is square and a fifth power

I have so far, for $n,y,q,p,z\in \mathbb{N}$

$n=30q$ , $n=y^2$ $\Rightarrow q=\frac{p^2}{30}$

And obviously $n=z^5$

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The answer is $30^{10}.$ Think about prime factorizations. –  Ragib Zaman Apr 22 '12 at 16:56

1 Answer 1

up vote 7 down vote accepted

Clearly $n=(2\cdot3\cdot5)^{10}$. If $n$ divisible by a prime $p$, and it's a square, it must be divisible by $p^2$. Similarly, if it's a fifth power, it must be divisible by $p^5$ and hence by $p^{10}$.

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Thank you very much! That is very helpful –  Freeman Apr 22 '12 at 17:06

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