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

Revision as of 12:33, 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.

$90-81=9=3^2$ , $76-44=32=2^5$ , and $1421-125=1296$ . So $p=2$ and $r=5$ . To find $s$ you need to memorize what is special about $1296$ , but if you haven't,

$6*6=36$

$6*36=216$

$216*6=1296=6^4$ .

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 <math>6*6=36</math>
 <math>6*36=216</math>
 <math>216*6=1296=6^4</math>.

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

Revision as of 12:33, 27 November 2013

Problem

Solution

See Also