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

(Solution)
m (Solution)
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<math>76-44=32=2^5</math>, so  <math>r=5</math>.   
 
<math>76-44=32=2^5</math>, so  <math>r=5</math>.   
  
Additionally, <math>1421-125=1296</math>.
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Next, <math>1421-125=1296</math>. Then, we find <math>s</math>.
 
 
So <math>p=2</math> and <math>r=5</math>. Then, we find <math>s</math>.
 
  
 
<math>6*6=36</math>
 
<math>6*6=36</math>
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<math>6*36=216</math>
 
<math>6*36=216</math>
  
<math>216*6=1296=6^4</math>.
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<math>6*216=1296=6^4</math>, so <math>s=4</math>.
  
 
It may help to memorize that <math>1296</math> is <math>6^4</math>.
 
It may help to memorize that <math>1296</math> is <math>6^4</math>.

Revision as of 23:53, 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.

$90-81=9=3^2$, so $p=2$.

$76-44=32=2^5$, so $r=5$.

Next, $1421-125=1296$. Then, we find $s$.

$6*6=36$

$6*36=216$

$6*216=1296=6^4$, so $s=4$.

It may help to memorize that $1296$ is $6^4$.

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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