Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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− | A | + | ==Problem== |
+ | What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>? | ||
+ | |||
+ | <math> \textbf{(A)}\ 2^2^4<10^8<5^1^2 </math> | ||
+ | <math> \textbf{(B)}\ 2^2^4<5^1^2<10^8 </math> | ||
+ | <math> \textbf{(C)}\ 5^1^2<2^2^4<10^8 </math> | ||
+ | <math> \textbf{(D)}\ 10^8<5^1^2<2^2^4</math> | ||
+ | <math> \textbf{(E)}\ 10^8<2^2^4<5^1^2 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | 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)}\ 2^2^4<10^8<5^1^2 }</math> is the correct answer. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2011|before=First Problem|num-a=2}} |
Revision as of 13:52, 4 March 2012
Problem
What is the correct ordering of the three numbers, , , and ?
$\textbf{(A)}\ 2^2^4<10^8<5^1^2$ (Error compiling LaTeX. ! Double superscript.) $\textbf{(B)}\ 2^2^4<5^1^2<10^8$ (Error compiling LaTeX. ! Double superscript.) $\textbf{(C)}\ 5^1^2<2^2^4<10^8$ (Error compiling LaTeX. ! Double superscript.) $\textbf{(D)}\ 10^8<5^1^2<2^2^4$ (Error compiling LaTeX. ! Double superscript.) $\textbf{(E)}\ 10^8<2^2^4<5^1^2$ (Error compiling LaTeX. ! Double superscript.)
Solution
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 $\boxed{\textbf{(A)}\ 2^2^4<10^8<5^1^2 }$ (Error compiling LaTeX. ! Double superscript.) is the correct answer.
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |