Find the minimum values of x, y, and z If:  
\begin{aligned}
\ {n} & = \ 2x ^2 \\
\ {n} & = \ 3y^3 \\
\ {n} & = \ 5z^5 \\
\end{aligned}
Find the minimum value for $x, y,$ and $z$.
Note: $n$ is the minimum value you can get by solving the set of equations.
 A: I take it that $n$ is a positive integer, as are $x$, $y$, and $z$. Then $n$ must be divisible by the primes $2$, $3$, and $5$. To minimize $n$, we allow no other primes.
Let $n=2^a3^b5^c$.  Since $n=2x^2$, $a$ must be odd, and $b$ and $c$ even.  This is because in the prime power factorization of a perfect square, each prime must occur to an even power. 
Since $n=3y^3$, $a$ must be divisible by $3$, and $b-1$ must be divisible by $3$, and $c$ must be divisible by $3$.  This is because in the prime power factorization of a perfect cube, each prime must occur to a power divisible by $3$.  
Since $n=5z^5$, $a$ must be divisible by $5$, and $b$ must be divisible by $5$, and $c-1$ must be divisible by $5$.
Now we look for the smallest non-negative integers $a$, $b$, and $c$ that satisfy our conditions.
Start with $a$. We must have $a$ odd, $a$ divisible by $3$ and by $5$. It is clear that the cheapest $a$ is $15$.  
We want $b$ to be even, and a multiple of $5$, and we want $b-1$ to be divisible by $3$. So $b$ is a multiple of $10$. Easily, $b=10$ satisfies the additional condition $b-1$ is divisible by $3$. 
Do something similar to find the cheapest $c$.
