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 $n=z^5$
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 $n=z^5$
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 $p^{10}$.