# 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$