2013 AMC 8 Problems/Problem 15: Difference between revisions
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==Solution== | ==Solution== | ||
This can be brute-forced. | |||
<math>90-81=9=3^2</math>, <math>76-44=32=2^5</math>, and <math>1421-125=1296</math>. | |||
So <math>p=2</math> and <math>r=5</math>. To find <math>s</math> you need to memorize what is special about <math>1296</math>, but if you haven't, | |||
<math>6*6=36</math> | |||
<math>6*36=216</math> | |||
<math>216*6=1296=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 08:27, 27 November 2013
Problem
If 
, 
, and 
, what is the product of p, r, and s?
Solution
This can be brute-forced.
, 
, and 
.
So 
and 
. To find 
you need to memorize what is special about 
, but if you haven't,
.
Therefore the answer is 
.
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
