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

(→‎Solution 1)
(→‎Solution 1)
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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>
−
<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>==
 
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.
 
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.

Revision as of 16:46, 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. $10^8=100,000,000$ $5^{12}=44,140,625$ $2^{24}=16,777,216$ 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 (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

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