Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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\textbf{(E)}\ 10^8<2^{24}<5^{12} </math> | \textbf{(E)}\ 10^8<2^{24}<5^{12} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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 <math>10^2=100< | + | Use brute force. |
| + | 10^8=100,000,000 | ||
| + | 5^{12}=44,140,625 | ||
| + | 2^{24}=16,777,216 | ||
| + | Therefore, <math> 2^{24}<10^8<5^{12} is the answer. </math>\boxed{A}<math> | ||
| + | == 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 </math>10^2=100<math>, </math>5^3=125<math>, and </math>2^6=64<math>. Since </math>64<100<125<math>, it follows that </math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=23|num-a=25}} | {{AMC8 box|year=2010|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:42, 17 November 2017
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution 1
Use brute force.
10^8=100,000,000
5^{12}=44,140,625
2^{24}=16,777,216
Therefore, 
\boxed{A}
10^2=100
5^3=125
2^6=64
64<100<125
\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.
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. 
