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

(Solution 1)
(Solution 1)
Line 10: Line 10:
 
==Solution 1==
 
==Solution 1==
 
Use brute force.
 
Use brute force.
<math>10^8=100,000,000</math>
+
<math>10^8=100,000,000</math>,
<math>5^{12}=44,140,625</math>
+
<math>5^{12}=44,140,625</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.
+
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer.
  
 
== Solution <math>2</math>==
 
== Solution <math>2</math>==

Revision as of 17:47, 6 January 2018

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. $10^8=100,000,000$, $5^{12}=44,140,625$, and $2^{24}=16,777,216$. Therefore, $\boxed{\text{(A)}2^{24}<10^8<5^{12}}$ is the answer.

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

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