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

(Solution: poor solution before.)
m (Solution)
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<math>3^p + 3^4 = 90\\
 
<math>3^p + 3^4 = 90\\
 
3^p + 81 = 90\\
 
3^p + 81 = 90\\
3^p = 9</math>, so <math>p = 2</math>.
+
3^p = 9</math>
 +
Therefore, <math>p = 2</math>.
  
 
<math>2^r + 44 = 76\\
 
<math>2^r + 44 = 76\\
2^r = 32</math>, so <math>r = 5</math>.
+
2^r = 32</math>
 +
Therefore, <math>r = 5</math>.
  
 
<math>5^3 + 6^s = 1421\\
 
<math>5^3 + 6^s = 1421\\

Revision as of 13:41, 29 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

$3^p + 3^4 = 90\\ 3^p + 81 = 90\\ 3^p = 9$ Therefore, $p = 2$.

$2^r + 44 = 76\\ 2^r = 32$ Therefore, $r = 5$.

$5^3 + 6^s = 1421\\ 125 + 6^s = 1296$.

To most people, it would not be immediately evident that $6^4 = 1296$, so we can multiply 6's until we get the desired number:

$6\cdot6=36$

$6\cdot36=216$

$6\cdot216=1296=6^4$, so $s=4$.

Therefore the answer is $2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}$.

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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