2013 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 12:40, 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 $s$ ?

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

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

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

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

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

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 ==Solution==
+<math>3^p + 3^4 = 90\\
+^p + 81 = 90\\
+^p = 9</math>, so <math>p = 2</math>.
-This can be brute-forced.
+<math>2^r + 44 = 76\\
+^r = 32</math>, so <math>r = 5</math>.
-<math>90-81=9=3^2</math>, so <math>p=2</math>.
+<math>5^3 + 6^s = 1421\\
++ 6^s = 1296</math>.
-<math>76-44=32=2^5</math>, so  <math>r=5</math>.
+To most people, it would not be immediately evident that <math>6^4 = 1296</math>, so we can multiply 6's until we get the desired number:
-Next, <math>1421-125=1296</math>. Then, we find <math>s</math>.
+<math>6\cdot6=36</math>
-<math>6*6=36</math>
+<math>6\cdot36=216</math>
-<math>6*36=216</math>
+<math>6\cdot216=1296=6^4</math>, so <math>s=4</math>.
-<math>6*216=1296=6^4</math>, so <math>s=4</math>.
+Therefore the answer is <math>2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}</math>.
-It may help to memorize that <math>1296</math> is <math>6^4</math>.
-Therefore the answer is <math>2*5*4=\boxed{\textbf{(B)}\ 40}</math>.
 ==See Also==
 {{AMC8 box|year=2013|num-b=14|num-a=16}}
 {{MAA Notice}}