Difference between revisions of "2013 AMC 8 Problems/Problem 15"
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3^p + 81 = 90\ | 3^p + 81 = 90\ | ||
3^p = 9</math> | 3^p = 9</math> | ||
+ | |||
Therefore, <math>p = 2</math>. | Therefore, <math>p = 2</math>. | ||
<math>2^r + 44 = 76\ | <math>2^r + 44 = 76\ | ||
2^r = 32</math> | 2^r = 32</math> | ||
+ | |||
Therefore, <math>r = 5</math>. | Therefore, <math>r = 5</math>. | ||
<math>5^3 + 6^s = 1421\ | <math>5^3 + 6^s = 1421\ | ||
− | 125 + 6^s = 1296</math> | + | 125 + 6^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: | 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: |
Revision as of 12:41, 29 November 2013
Problem
If , , and , what is the product of , , and ?
Solution
Therefore, .
Therefore, .
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.