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

(Solution)
(Problem)
Line 4: Line 4:
 
<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\
 
<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\
 
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \
 
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \
  \textbf{(C)}\ 5^{1^2}<2^{24}<10^8 \
+
  \textbf{(C)}\ 5^{12}<2^{24}<10^8 \
 
\textbf{(D)}\ 10^8<5^{12}<2^{24} \
 
\textbf{(D)}\ 10^8<5^{12}<2^{24} \
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>

Revision as of 16:07, 18 March 2015

Problem

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

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

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 $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png