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 ![$2$](//latex.artofproblemsolving.com/4/1/c/41c544263a265ff15498ee45f7392c5f86c6d151.png)
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.