<s>[INST] find $n$ so $n/k$ is a $k$th power, $k=2,3,5$. Find a natural number n, in canonical form, such that:  
$n/2=a^2$   
$n/3=b^3$   
$n/5= c^5$  
for some a,b and c (natural numbers). 
 [/INST] Attempt an $n$ of the form $n = 2^x 3^y 5^z$.
You have $n/2 = 2^{x-1} 3^y 5^z$ as a square, so $x \equiv 1 (\bmod 2),\;  y \equiv z \equiv 0 (\bmod 2).$
Analogously, $x \equiv z \equiv 0 (\bmod 3)$, $y \equiv 1 (\bmod 3)$ and 
$x \equiv y \equiv 0 (\bmod 5)$, $z \equiv 1 (\bmod 5)$.
The Chinese remainder theorem guarantees that you will find such natural numbers $x,y,z$.
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