# 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. $10^8=100,000,000$, $5^{12}=244,140,625$, and $2^{24}=16,777,216$. Therefore, $\boxed{\text{(A)}2^{24}<10^8<5^{12}}$ is the answer. (Not recommended for this contest)

## 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.

## Solution 3

$First$, $let$ $us$ make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. $10^8$ is fine as is. We can rewrite $2^{24}$ as $(2^3)^8=8^8$. We can rewrite $5^{12}$ as $(5^{\frac{3}{2}})^8=(\sqrt{125})^8)$. We take the eighth root of all of these to get ${10, 8, \sqrt{125}}$. Obviously, $8<10<\sqrt{125}$, so the answer is $\textbf{(A)}\ 2^{24}<10^8<5^{12}$. $Solution$ $by$ $MathHayden$

## 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

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

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