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

(Problem)
(Solution)
Line 8: Line 8:
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
 
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math>
  
==Solution==
+
==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 <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.
+
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==
 +
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.
  
 
==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 01:42, 17 November 2017

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 2^{24}=16,777,216 Therefore, $2^{24}<10^8<5^{12} is the answer.$\boxed{A}$== 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.

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

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

Invalid username
Login to AoPS