2011 AMC 8 Problems/Problem 17: Difference between revisions

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

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

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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			==Problem==
	Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?		Let <math>w</math>, <math>x</math>, <math>y</math>, and <math>z</math> be whole numbers. If <math>2^w \cdot 3^x \cdot 5^y \cdot 7^z = 588</math>, then what does <math>2w + 3x + 5y + 7z</math> equal?