Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n=z^5$

share|cite|improve this question
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 $p^{10}$.

share|cite|improve this answer
Thank you very much! That is very helpful – Freeman Apr 22 '12 at 17:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.