Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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==Solution== | ==Solution== | ||
+ | <math>3^p + 3^4 = 90\ | ||
+ | 3^p + 81 = 90\ | ||
+ | 3^p = 9</math>, so <math>p = 2</math>. | ||
− | + | <math>2^r + 44 = 76\ | |
+ | 2^r = 32</math>, so <math>r = 5</math>. | ||
− | <math> | + | <math>5^3 + 6^s = 1421\ |
+ | 125 + 6^s = 1296</math>. | ||
− | <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 | + | <math>6\cdot36=216</math> |
− | <math>6 | + | <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>. | |
− | |||
− | |||
− | |||
− | Therefore the answer is <math>2 | ||
==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 , , and , what is the product of , , and ?
Solution
, so .
, so .
.
To most people, it would not be immediately evident that , so we can multiply 6's until we get the desired number:
, so .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.