Difference between revisions of "2010 AMC 8 Problems/Problem 24"
m (change of italics) |
m (→Solution 3) |
||
Line 20: | Line 20: | ||
== Solution 3== | == Solution 3== | ||
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. | First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. | ||
− | <math>10^8</math> is fine as is. | + | <math>10^8</math> is as fine as it is. |
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>. | We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>. | ||
− | + | Then 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>\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 MathHayden | + | <math>Solution by 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 15:19, 2 August 2020
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 as fine as it is. We can rewrite as . Then we can rewrite as . We take the eighth root of all of these to get . Obviously, , so the answer is .
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.