Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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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 | + | == 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 <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. | ||
+ | |||
+ | == Solution 3== | ||
+ | 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. | ||
+ | We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</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>. | ||
+ | Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>. | ||
+ | Solution by coolak | ||
==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 18:09, 6 January 2018
Problem
What is the correct ordering of the three numbers, , , and ?
Solution 1
Use brute force. , , and . Therefore, is the answer.
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 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.