Use the Chinese Remainder Theorem to find the smallest element of $$\left\{n\in \mathbb{N}: \sqrt{\frac{n}{2}}, \sqrt[3]{\frac{n}{3}}, \sqrt[5]{\frac{n}{5}}\in \mathbb{N}\right\}$$

I have played around it and I have obtained the set of congruences

$$ n\equiv 0 \mod{2k_{1}^{2}}$$ $$ n\equiv 0 \mod{3k_{2}^{3}}$$ $$ n\equiv 0 \mod{5k_{3}^{5}}$$ for some $k_{1}, k_{2}, k_{3}\in \mathbb{N}$. But since they might not be mutually coprime I'm not sure how the Chinese Remainder Theorem can be applied.


Ok. At least it is clear that our number has to be of the form $2^x3^y5^z$.All we have to do is determine the least $x,y,z$ so that our conditions are satisfied.

i) Note that$2^{x-1}3^y5^z$ is a square number. Hence 2 divides $x-1$, $y$ and $z$.

ii) Note that $2^x3^{y-1}5^z$ is a perfect cube. Hence 3 divides $x$, $y-1$ and $z$.

iii) Note that $2^x3^{y}5^{z-1}$ is a perfect fifth power. Hence 5 divides $x$,$y$ and $z-1$.

So you have the following:

i) $x \equiv 1(2), x\equiv 0(3),x\equiv 0(5)$

ii)$y \equiv 0(2), y\equiv 1(3),y\equiv 0(5)$

iii)$z \equiv 0(2), x\equiv 0(3),x\equiv 1(5)$

Solve for $x,y,z$ using Chinese Remainder theorem. That is your job done. I expect the answer then, to be $2^{15}3^{10}5^{6} = 30233088000000$.


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