Difference between revisions of "2010 AMC 8 Problems/Problem 24"

(Solution 1)
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Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer.
 
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer.
  
== Solution <math>2</math>==
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== 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 <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
 
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 <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
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 +
== Solution 3==
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First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
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<math>10^8</math> is fine as is.
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We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
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We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>.
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We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.
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Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.
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Solution by coolak
  
 
==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 18:09, 6 January 2018

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.

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 coolak

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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