Difference between revisions of "2022 AMC 10B Problems/Problem 5"
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− | + | <cmath>\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} | |
= \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{2^{10} \cdot 3^{2}}{3^2\cdot 5^2\cdot 7^2}}} | = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{2^{10} \cdot 3^{2}}{3^2\cdot 5^2\cdot 7^2}}} | ||
= \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\frac{2^5 \cdot 3}{3\cdot5\cdot 7}} | = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\frac{2^5 \cdot 3}{3\cdot5\cdot 7}} | ||
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~TaeKim | ~TaeKim | ||
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+ | ~minor edits by mathboy100 | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 12:35, 28 November 2022
Contents
Problem
What is the value of
Solution 1
We apply the difference of squares to the denominator, and then regroup factors:
~MRENTHUSIASM
Solution 2 (Brute Force)
Since these numbers are fairly small, we can use brute force as follows:
~not_slay
Solution 2.1 (Brute Force + Cancellation)
This solution starts off exactly as the one above. We simplify to get:
But now, we can get a nice simplification as shown:
~TaeKim
~minor edits by mathboy100
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 AMC 10 Problems and Solutions |
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.