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

m
Line 4: Line 4:
  
 
==Solution==
 
==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==
 
==See Also==
 
{{AMC8 box|year=2011|num-b=16|num-a=18}}
 
{{AMC8 box|year=2011|num-b=16|num-a=18}}

Revision as of 19:30, 25 November 2011

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
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