Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 8 Problems/Problem 24"
(→‎Solution 3)
 
(33 intermediate revisions by 17 users not shown)
Line 8: Line 8:
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
  
−
==Solution 1==
+
== Solution 1==
−
Use brute force.
+
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.
−
<math>10^8=100,000,000</math>,  
 
−
<math>5^{12}=44,140,625</math>, and
 
−
<math>2^{24}=16,777,216</math>.
 
−
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer.
 
  
 
== Solution 2==
 
== 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.
 
−
 
−
== 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.
+
<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>.
+
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>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math>.
−
Solution by coolak
+
Solution by Math
 +
 
 +
==Solution 3==
 +
We know that <math>10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}</math>. We also know that <math>5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}</math>. If we remove the common factor of <math>5^{8}</math> from both expressions, we are left with <math>2^{8}</math>, which equals 256, and <math>5^{4}</math>, which equals 625. So we know that <math>5^{12}</math> is bigger than <math>10^{8}</math>. Now we need to figure out which is bigger, <math>10^{8}</math> or <math>2^{24}</math>. To do this, we rewrite <math>2^{24}</math> as <math>(2^{3})^{8}=(8)^{8}</math>, which is clearly less than <math>10^{8}</math>. Therefore, <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
 +
 
 +
By naman14
 +
 
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/rQUwNC0gqdg?t=381
 +
 
 +
==Video by MathTalks==
 +
 
 +
https://youtu.be/mSCQzmfdX-g
 +
 
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/EfCyJF1FEO
 +
~someone
  
 
==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 07:12, 9 November 2024

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

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 2

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 $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$. Solution by Math

Solution 3

We know that $10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}$. We also know that $5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}$. If we remove the common factor of $5^{8}$ from both expressions, we are left with $2^{8}$, which equals 256, and $5^{4}$, which equals 625. So we know that $5^{12}$ is bigger than $10^{8}$. Now we need to figure out which is bigger, $10^{8}$ or $2^{24}$. To do this, we rewrite $2^{24}$ as $(2^{3})^{8}=(8)^{8}$, which is clearly less than $10^{8}$. Therefore, $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

By naman14

Video Solution by OmegaLearn

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

Video by MathTalks

https://youtu.be/mSCQzmfdX-g


Video Solution by WhyMath

https://youtu.be/EfCyJF1FEO ~someone

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_24&oldid=232529"