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

m (Solution)
(Solution: poor solution before.)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
 +
<math>3^p + 3^4 = 90\
 +
3^p + 81 = 90\
 +
3^p = 9</math>, so <math>p = 2</math>.
  
This can be brute-forced.
+
<math>2^r + 44 = 76\
 +
2^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\
 +
125 + 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==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=14|num-a=16}}
 
{{AMC8 box|year=2013|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:40, 29 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

$3^p + 3^4 = 90\\ 3^p + 81 = 90\\ 3^p = 9$, 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 $2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}$.

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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

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