2013 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 12:41, 29 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 $s$ ?

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

$3^p + 3^4 = 90\\ 3^p + 81 = 90\\ 3^p = 9$ Therefore, $p = 2$ .

$2^r + 44 = 76\\ 2^r = 32$ Therefore, $r = 5$ .

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

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
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 7: / Line 7: @@
 <math>3^p + 3^4 = 90\\
 ^p + 81 = 90\\
-^p = 9</math>, so <math>p = 2</math>.
+^p = 9</math>
+Therefore, <math>p = 2</math>.
 <math>2^r + 44 = 76\\
-^r = 32</math>, so <math>r = 5</math>.
+^r = 32</math>
+Therefore, <math>r = 5</math>.
 <math>5^3 + 6^s = 1421\\