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2010 AMC 8 Problems/Problem 24: Difference between revisions

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== Solution 3==
== Solution 3==
<math>First</math>, <math>let</math> <math>us</math> make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
<math>10^8</math> is fine as is.
<math>10^8</math> is fine as is.
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
Line 25: Line 25:
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.
<math>Solution</math> <math>by</math> <math>MathHayden</math>
Solution by MathHayden


==See Also==
==See Also==
{{AMC8 box|year=2010|num-b=23|num-a=25}}
{{AMC8 box|year=2010|num-b=23|num-a=25}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 14:17, 2 August 2020

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 by
Problem 23 		Followed by
Problem 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

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