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

Revision as of 15:22, 31 December 2022

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 the contest and will take forever)

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

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=381

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

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

@@ Line 28: / Line 28: @@
 Solution by MathHayden
-==Video Solution==
+==Video Solution by OmegaLearn==
 https://youtu.be/rQUwNC0gqdg?t=381
 ==See Also==
 {{AMC8 box|year=2010|num-b=23|num-a=25}}
 {{MAA Notice}}

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