2010 AMC 8 Problems/Problem 24: Difference between revisions
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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 | == 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 <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. | ||
== Solution 3== | |||
First, let us 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. | |||
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 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>. | |||
Solution by coolak | |||
==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 17:09, 6 January 2018
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution 1
Use brute force.
,
, and
.
Therefore, 
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 
, 
, and 
. Since 
, it follows that 
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.
is fine as is.
We can rewrite as 
.
We can rewrite as 
.
We take the eighth root of all of these to get 
.
Obviously, 
, so the answer is 
.
Solution by coolak
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
