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. (Definitely recommended for the contest) |
== Solution 2== | == Solution 2== |
Revision as of 19:55, 8 October 2020
Problem
What is the correct ordering of the three numbers, , , and ?
Solution 1
Use brute force. , , and . Therefore, is the answer. (Definitely recommended for the 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 as fine as it is. We can rewrite as . Then we can rewrite as . We take the eighth root of all of these to get . Obviously, , so the answer is . Solution by MathHayden
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.