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. $10^8=100,000,000$ , $5^{12}=44,140,625$ , and $2^{24}=16,777,216$ . Therefore, $\boxed{\text{(A)}2^{24}<10^8<5^{12}}$ is the answer.

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.

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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2010 AMC 8 Problems/Problem 24

Contents

Problem

Solution 1

Solution $2$

See Also