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

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This can be brute-forced.
 
This can be brute-forced.
  
<math>90-81=9=3^2</math>, <math>76-44=32=2^5</math>, and <math>1421-125=1296</math>.
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<math>90-81=9=3^2</math>,  
 +
 
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<math>76-44=32=2^5</math>, and  
 +
 
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<math>1421-125=1296</math>.
 +
 
 
So <math>p=2</math> and <math>r=5</math>. To find <math>s</math> you need to memorize what is special about <math>1296</math>, but if you haven't,
 
So <math>p=2</math> and <math>r=5</math>. To find <math>s</math> you need to memorize what is special about <math>1296</math>, but if you haven't,
  

Revision as of 19:33, 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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