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

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What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?
 
What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?
  
<math> \textbf{(A)}\ 2^2^4<10^8<5^1^2 </math>
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<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\\
<math> \textbf{(B)}\ 2^2^4<5^1^2<10^8 </math>
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\textbf{(B)}\ 2^{24}<5^{12}<10^8 \\
<math> \textbf{(C)}\ 5^1^2<2^2^4<10^8 </math>
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\textbf{(C)}\ 5^{12}<2^{24}<10^8 \\
<math> \textbf{(D)}\ 10^8<5^1^2<2^2^4</math>
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\textbf{(D)}\ 10^8<5^{12}<2^{24} \\
<math> \textbf{(E)}\ 10^8<2^2^4<5^1^2 </math>
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\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
  
==Solution==
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==Video Solution==
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)}\ 2^2^4<10^8<5^1^2 }</math> is the correct answer.
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https://youtu.be/rQUwNC0gqdg?t=381
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==Solution 1==
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Use brute force.
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<math>10^8=100,000,000</math>,
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<math>5^{12}=244,140,625</math>, and
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<math>2^{24}=16,777,216</math>.
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Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for the contest and will take forever)
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== Solution 2==
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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 as fine as it is.
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We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
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Then 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 MathHayden
  
 
==See Also==
 
==See Also==
{{AMC8 box|year=2011|before=First Problem|num-a=2}}
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{{AMC8 box|year=2010|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 19:37, 7 March 2021

Problem

What is the correct ordering of the three numbers, $10^8$, $5^{12}$, and $2^{24}$?

$\textbf{(A)}\ 2^{24}<10^8<5^{12}\\ \textbf{(B)}\ 2^{24}<5^{12}<10^8 \\  \textbf{(C)}\ 5^{12}<2^{24}<10^8 \\ \textbf{(D)}\ 10^8<5^{12}<2^{24} \\ \textbf{(E)}\ 10^8<2^{24}<5^{12}$

Video Solution

https://youtu.be/rQUwNC0gqdg?t=381

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 the contest and will take forever)

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 as fine as it is. We can rewrite $2^{24}$ as $(2^3)^8=8^8$. Then 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 (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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