Difference between revisions of "2011 AMC 8 Problems/Problem 17"

Revision as of 02:03, 5 July 2013

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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?
-<math> \text{(A) } 21\qquad\text{(B) }25\qquad\text{(C) }27\qquad\text{(D) }35\qquad\text{(E) }56 </math>
+<math> \textbf{(A) } 21\qquad\textbf{(B) }25\qquad\textbf{(C) }27\qquad\textbf{(D) }35\qquad\textbf{(E) }56 </math>
 ==Solution==
+The [[prime factorization]] of <math>588</math> is <math>2^2\cdot3\cdot7^2.</math> We can see <math>w=2, x=1,</math> and <math>z=2.</math> Because <math>5^0=1, y=0.</math>
+<cmath>2w+3x+5y+7z=4+3+0+14=\boxed{\textbf{(A)}\ 21}</cmath>
 ==See Also==
 {{AMC8 box|year=2011|num-b=16|num-a=18}}
+{{MAA Notice}}