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

(Problem)
(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>?
 
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^{24}<10^8<5^{12}</math>
+
<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\\
<math> \textbf{(B)}\ 2^{24}<5^{12}<10^8 </math>
+
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \\
<math> \textbf{(C)}\ 5^{1^2}<2^{24}<10^8 </math>
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\textbf{(C)}\ 5^{1^2}<2^{24}<10^8 \\
<math> \textbf{(D)}\ 10^8<5^{12}<2^{24}</math>
+
\textbf{(D)}\ 10^8<5^{12}<2^{24} \\
<math> \textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
+
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
  
 
==Solution==
 
==Solution==

Revision as of 17:06, 18 March 2015

Problem

What is the correct ordering of the three numbers, $10^8$, $5^{12}$, and $2^{24}$?

$\textbf{(A)}\ 2^{24}<10^8<5^{12}\\ \textbf{(B)}\ 2^{24}<5^{12}<10^8 \\  \textbf{(C)}\ 5^{1^2}<2^{24}<10^8 \\ \textbf{(D)}\ 10^8<5^{12}<2^{24} \\ \textbf{(E)}\ 10^8<2^{24}<5^{12}$

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

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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