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

Revision as of 14:34, 5 November 2017

$6\cdot6=36$

Solve for s, x, and r.

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 1: / Line 1: @@
-==Solution==
+==Problem==
-<math>3^p + 3^4 = 90\\
-^p + 81 = 90\\
-^p = 9</math>
-Therefore, <math>p = 2</math>.
-<math>2^r + 44 = 76\\
-^r = 32</math>
-Therefore, <math>r = 5</math>.
-<math>5^3 + 6^s = 1421\\
-+ 6^s = 1421\\
-^s=1296</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:
 <math>6\cdot6=36</math>
-<math>6\cdot36=216</math>
+Solve for s, x, and r.
-<math>6\cdot216=1296=6^4</math>, so <math>s=4</math>.
-Therefore the answer is <math>2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}</math>.
 ==See Also==
 {{AMC8 box|year=2013|num-b=14|num-a=16}}
 {{MAA Notice}}