Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>. | We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>. | ||
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>. | We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>. | ||
| − | Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\ | + | Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textzf{(A)}\ 2^{24}<10^8<5^{12}</math>. |
Solution by coolak | Solution by coolak | ||
Revision as of 20:53, 7 November 2019
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution 1
Use brute force.
,
, and
.
Therefore, 
is the answer. (Not recommended for this contest)
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 
, 
, and 
. Since 
, it follows that 
is the correct answer.
Solution 3
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
is fine as is.
We can rewrite as 
.
We can rewrite as 
.
We take the eighth root of all of these to get 
.
Obviously, 
, so the answer is $\textzf{(A)}\ 2^{24}<10^8<5^{12}$ (Error compiling LaTeX. Unknown error_msg).
Solution by coolak
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. 
