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

(→‎Solution 3)
(→‎Solution 3)
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== Solution 3==
 
== Solution 3==
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First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
+
<math>First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
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<math>10^8</math> is fine as is.
+
</math>10^8<math> is fine as is.
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We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
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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>.
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We can rewrite </math>5^{12}<math> as </math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)<math>.
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We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.
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We take the eighth root of all of these to get </math>{10, 8, \sqrt{125}}<math>.
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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>\textbf{(A)}\ 2^{24}<10^8<5^{12}<math>.
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Solution by <math>MathHayden</math>
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Solution by MathHayden</math>
  
 
==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 20:55, 7 November 2019

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. $10^8=100,000,000$, $5^{12}=244,140,625$, and $2^{24}=16,777,216$. Therefore, $\boxed{\text{(A)}2^{24}<10^8<5^{12}}$ is the answer. (Not recommended for this contest)

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 fine as is. We can rewrite$2^{24}$as$(2^3)^8=8^8$. 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$\textbf{(A)}\ 2^{24}<10^8<5^{12}$. Solution by MathHayden$

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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