2011 AMC 8 Problems/Problem 17

Problem

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?

$\textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56$

Solution

The prime factorization of $588$ is $2^2\cdot3\cdot7^2.$ We can see $w=2, x=1,$ and $z=2.$ Because $5^0=1, y=0.$

\[2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}\]

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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