Let $w$, $x$, $y$, and $z$ be whole numbers. If $2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588$, then what does $2w + 3x + 5y + 7z$ equal?
$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$
The [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization) of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$
\[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]