Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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<math>5^{12}=244,140,625</math>, and | <math>5^{12}=244,140,625</math>, and | ||
<math>2^{24}=16,777,216</math>. | <math>2^{24}=16,777,216</math>. | ||
− | 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. (Highly recommended for the contest and will take no time at all trust) |
== Solution 2== | == Solution 2== |
Revision as of 17:46, 1 July 2024
Contents
[hide]Problem
What is the correct ordering of the three numbers, , , and ?
Solution 1
Use brute force. , , and . Therefore, is the answer. (Highly recommended for the contest and will take no time at all trust)
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 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 . This leaves us with only 2 options, A or E. 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.