Difference between revisions of "2010 AMC 8 Problems/Problem 24"
Mathhayden (talk | contribs) (→Solution 3) |
Mathhayden (talk | contribs) (→Solution 3) |
||
Line 19: | Line 19: | ||
== 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. | + | <math>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>10^8<math> is fine as is. |
− | We can rewrite <math>2^{24}< | + | We can rewrite </math>2^{24}<math> as </math>(2^3)^8=8^8<math>. |
− | We can rewrite <math>5^{12}< | + | 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}}< | + | We take the eighth root of all of these to get </math>{10, 8, \sqrt{125}}<math>. |
− | Obviously, <math>8<10<\sqrt{125}< | + | 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 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 21:55, 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
10^82^{24}(2^3)^8=8^85^{12}(5^{\frac{3}{2}})^8=(\sqrt{125})^8){10, 8, \sqrt{125}}8<10<\sqrt{125}\textbf{(A)}\ 2^{24}<10^8<5^{12}
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.