2013 AMC 8 Problems/Problem 15

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

First, we're going to solve for $p$ . Start with $3^p+$ 3^4=90 $. Then, change 3^4 to$ 81 $. Subtract$ 81 $from both sides to get$ 3^p=9 $and see that$ p $is$ 2 $. Now, solve for$ r $. Since$ 2^r+44=76 $,$ 2^r $must equal$ 32 $, so$ r=5 $. Now, solve for$ s $.$ 5^3+6^s=1421 $can be simplified to$ 125+6^s=1421 $which simplifies further to$ 6^s=1296 $. Therefore,$ s=4 $.$ prs $equals$ 2*5*4 $which equals$ 40 $. So, the answer is$ \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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2013 AMC 8 Problems/Problem 15

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