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

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\textbf{(D)}\ 10^8<5^{12}<2^{24} \\
 
\textbf{(D)}\ 10^8<5^{12}<2^{24} \\
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
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==Video Solution==
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https://youtu.be/rQUwNC0gqdg?t=381
  
 
==Solution 1==
 
==Solution 1==
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<math>5^{12}=244,140,625</math>, and
 
<math>5^{12}=244,140,625</math>, and
 
<math>2^{24}=16,777,216</math>.
 
<math>2^{24}=16,777,216</math>.
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer. (Not recommended for this contest)
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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)
  
 
== Solution 2==
 
== Solution 2==
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== Solution 3==
 
== Solution 3==
 
First, let us 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.
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<math>10^8</math> is as fine as it 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>.
We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^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>.
 
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>.
Solution by <math>MathHayden</math>
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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}}

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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