Difference between revisions of "2010 AMC 8 Problems/Problem 24"
(→Solution 1) |
(→Solution 1) |
||
Line 11: | Line 11: | ||
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> |
− | <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>== | ||
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. | 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. |
Revision as of 17:46, 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.