2010 AMC 8 Problems/Problem 24: Difference between revisions

Revision as of 20:56, 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 $\textbf{(A)}\ 2^{24}<10^8<5^{12}$ . $Solution$ $by$ $MathHayden$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 19: / Line 19: @@
 == Solution 3==
-<math>First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
+<math>First</math>, <math>let</math> <math>us</math> make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
-</math>10^8<math> is fine as is.
+<math>10^8</math> is fine as is.
-We can rewrite </math>2^{24}<math> as </math>(2^3)^8=8^8<math>.
+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>.
+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>.
+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>\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>.
-Solution by MathHayden</math>
+<math>Solution</math> <math>by</math> <math>MathHayden</math>
 ==See Also==
 {{AMC8 box|year=2010|num-b=23|num-a=25}}
 {{MAA Notice}}

Page

Toolbox

Search