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

## 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 1

Use brute force. 10^8=100,000,000 5^{12}=44,140,625 2^{24}=16,777,216 Therefore, $2^{24}<10^8<5^{12} is the answer.$\boxed{A}$== 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$10^2=100$,$5^3=125$, 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 (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 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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