Difference between revisions of "2022 AMC 10B Problems/Problem 5"
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==Solution 3 (Brute Force)== | ==Solution 3 (Brute Force)== | ||
− | This solution starts | + | This solution starts precisely like the one above. We simplify to get the following: |
<cmath>\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \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)}}}</cmath> | <cmath>\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \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)}}}</cmath> | ||
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~minor edits by mathboy100 | ~minor edits by mathboy100 | ||
− | ==Video Solution | + | ==Video Solution (⚡️2 min solution⚡️)== |
https://youtu.be/N7hGuy0MWOQ | https://youtu.be/N7hGuy0MWOQ | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/_KNR0JV5rdI?t=470 | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2022|ab=B|num-b=4|num-a=6}} | ||
− | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 19:59, 8 September 2023
Contents
Problem
What is the value of
Solution 1 (Difference of Squares)
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 3 (Brute Force)
This solution starts precisely like the one above. We simplify to get the following:
But now, we can get a nice simplification as shown:
~TaeKim
~minor edits by mathboy100
Video Solution (⚡️2 min solution⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=470
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.