Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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<math>5^{12}=244,140,625</math>, and | <math>5^{12}=244,140,625</math>, and | ||
<math>2^{24}=16,777,216</math>. | <math>2^{24}=16,777,216</math>. | ||
− | Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. | + | Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for this contest) |
== Solution 2== | == Solution 2== |
Revision as of 17:36, 5 October 2019
Problem
What is the correct ordering of the three numbers, , , and ?
Solution 1
Use brute force. , , and . Therefore, is the answer. (Not recommended for this contest)
Solution 2
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get , , and . Since , it follows that is the correct answer.
Solution 3
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. is fine as is. We can rewrite as . We can rewrite as . We take the eighth root of all of these to get . Obviously, , so the answer is . Solution by coolak
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.