2013 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 14:33, 5 November 2017

$3^p + 3^4 = 90\\ 3^p + 81 = 90\\ 3^p = 9$

Therefore, $p = 2$ .

$2^r + 44 = 76\\ 2^r = 32$

Therefore, $r = 5$ .

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

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 $2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}$ .

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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-==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 <math>s</math>?
-<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math>
 ==Solution==
 <math>3^p + 3^4 = 90\\