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

Revision as of 23:51, 27 November 2013

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 $r=5$ .

Additionally, $1421-125=1296$ .

So $p=2$ and $r=5$ . Then, we find $s$ .

$6*6=36$

$6*36=216$

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

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)
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

@@ Line 8: / Line 8: @@
 This can be brute-forced.
-<math>90-81=9=3^2</math>,
+<math>90-81=9=3^2</math>, so <math>p=2</math>.
-<math>76-44=32=2^5</math>, and
+<math>76-44=32=2^5</math>, so  <math>r=5</math>.
-<math>1421-125=1296</math>.
+Additionally, <math>1421-125=1296</math>.
 So <math>p=2</math> and <math>r=5</math>. Then, we find <math>s</math>.