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)
(→‎Solution 1)
(32 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==
 
−
Use brute force.
 
−
10^8=100,000,000
 
−
5^{12}=44,140,625
 
−
2^{24}=16,777,216
 
−
Therefore, <math> 2^{24}<10^8<5^{12} is the answer. </math>\boxed{A}<math>
 
 
== 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}}$ 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.
 +
 
 +
== Solution 3==
 +
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 as fine as it is.
 +
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</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>.
 +
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math>.
 +
Solution by Math
 +
 
 +
==Solution 4==
 +
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>. This leaves us with only 2 options, A or E. 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^{8})^{3}=(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}}

Revision as of 10:09, 15 July 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 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 $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$. Solution by Math

Solution 4

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 $10^{8}$. This leaves us with only 2 options, A or E. Now we need to figure out which is bigger, $10^{8}$ or $2^{24}$. To do this, we rewrite $2^{24}$ as $(2^{8})^{3}=(2^{3})^{8}=(8)^{8}$, 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=222596"