Difference between revisions of "2013 AMC 8 Problems/Problem 15"

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==Problem==
 
==Problem==
If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and s?
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If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>?
  
 
<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math>
 
<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math>

Revision as of 13:32, 27 November 2013

Problem

If $3^p + 3^4 = 90$, $2^r + 44 = 76$, and $5^3 + 6^s = 1421$, what is the product of $p$, $r$, and $s$?

$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$

Solution

This can be brute-forced.

$90-81=9=3^2$, $76-44=32=2^5$, and $1421-125=1296$. So $p=2$ and $r=5$. To find $s$ you need to memorize what is special about $1296$, but if you haven't,

$6*6=36$ $6*36=216$ $216*6=1296=6^4$.

Therefore the answer is $2*5*4=\boxed{\textbf{(B)}\ 40}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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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