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 $s$ ?

$\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90$

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 $2*5*4=\boxed{\textbf{(B)}\ 40}$ .

2013 AMC 8 (Problems • Answer Key • Resources)
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