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} $. Solution by MathHayden$

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

Solution 3

See Also