Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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\textbf{(E)}\ 10^8<2^{24}<5^{12} </math> | \textbf{(E)}\ 10^8<2^{24}<5^{12} </math> | ||
− | ==Solution 1== | + | == Solution 1== |
− | + | 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. | |
− | <math>10^ | ||
− | <math>5^ | ||
− | <math>2^ | ||
− | |||
== Solution 2== | == Solution 2== | ||
− | |||
− | |||
− | |||
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. | 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. | + | <math>10^8</math> is as fine as it 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>. | ||
− | + | Then 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>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math>. |
− | Solution by | + | Solution by Math |
+ | |||
+ | ==Solution 4== | ||
+ | We know that <math>10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}</math>. We also know that <math>5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}</math>. If we remove the common factor of <math>5^{8}</math> from both expressions, we are left with <math>2^{8}</math>, which equals 256, and <math>5^{4}</math>, which equals 625. So we know that <math>5^{12}</math> is bigger than <math>10^{8}</math>. Now we need to figure out which is bigger, <math>10^{8}</math> or <math>2^{24}</math>. To do this, we rewrite <math>2^{24}</math> as <math>(2^{3})^{8}=(8)^{8}</math>, which is clearly less than <math>10^{8}</math>. Therefore, <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer. | ||
+ | |||
+ | By naman14 | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=381 | ||
+ | |||
+ | ==Video by MathTalks== | ||
+ | |||
+ | https://youtu.be/mSCQzmfdX-g | ||
+ | |||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/EfCyJF1FEO | ||
+ | ~someone | ||
==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}} |
Latest revision as of 20:59, 23 October 2024
Contents
Problem
What is the correct ordering of the three numbers, , , and ?
Solution 1
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 2
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. is as fine as it is. We can rewrite as . Then we can rewrite as . We take the eighth root of all of these to get . Obviously, , so the answer is . Solution by Math
Solution 4
We know that . We also know that . If we remove the common factor of from both expressions, we are left with , which equals 256, and , which equals 625. So we know that is bigger than . Now we need to figure out which is bigger, or . To do this, we rewrite as , which is clearly less than . Therefore, is the correct answer.
By naman14
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=381
Video by MathTalks
Video Solution by WhyMath
https://youtu.be/EfCyJF1FEO ~someone
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.