Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
Use brute force. | Use brute force. | ||
| − | 10^8=100,000,000 | + | <math>10^8=100,000,000</math> |
| − | 5^ | + | <math>5^12=44,140,625</math> |
| − | 2^ | + | <math>2^24=16,777,216</math> |
| − | Therefore, <math> 2^ | + | Therefore, <math>\boxed{\text{(A)}2^24<10^8<5^12}</math> is the answer. |
| − | + | == Solution <math>2</math>== | |
| − | 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 < | + | 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}}</math> 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 16:45, 6 January 2018
Contents
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution 1
Use brute force.
Therefore, 
is the answer.
Solution 
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.
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. 
