Difference between revisions of "2010 AMC 8 Problems/Problem 24"

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==Problem==
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What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?
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<math> \textbf{(A)}\ 2^2^4<10^8<5^1^2 </math>
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<math> \textbf{(B)}\ 2^2^4<5^1^2<10^8 </math>
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<math> \textbf{(C)}\ 5^1^2<2^2^4<10^8 </math>
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<math> \textbf{(D)}\ 10^8<5^1^2<2^2^4</math>
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<math> \textbf{(E)}\ 10^8<2^2^4<5^1^2 </math>
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==Solution==
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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.
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==See Also==
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{{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, $10^8$, $5^{12}$, and $2^{24}$?

$\textbf{(A)}\ 2^2^4<10^8<5^1^2$ (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 $10^2=100$, $5^3=125$, and $2^6=64$. Since $64<100<125$, 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 (ProblemsAnswer KeyResources)
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
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