2010 AMC 8 Problems/Problem 24: Difference between revisions
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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>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>. | Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>. | ||
Solution by | Solution by <math>MathHayden</math> | ||
==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 20:54, 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 
.
Solution by 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
