Difference between revisions of "2010 AMC 8 Problems/Problem 24"
(→Solution 1) |
(→Solution 1) |
||
| Line 10: | Line 10: | ||
==Solution 1== | ==Solution 1== | ||
Use brute force. | Use brute force. | ||
| − | <math>10^8=100,000,000</math> | + | <math>10^8=100,000,000</math>, |
| − | <math>5^{12}=44,140,625</math> | + | <math>5^{12}=44,140,625</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. |
== Solution <math>2</math>== | == Solution <math>2</math>== | ||
Revision as of 17:47, 6 January 2018
Contents
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution 1
Use brute force.
,
, and
.
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. 
