Difference between revisions of "2022 AMC 10B Problems/Problem 5"
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=\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{\textbf{(B)}\ 2}.</cmath> | =\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{\textbf{(B)}\ 2}.</cmath> | ||
~not_slay | ~not_slay | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/N7hGuy0MWOQ | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See Also == | == See Also == | ||
Revision as of 19:41, 19 November 2022
Contents
Problem
What is the value of ![\[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\]](http://latex.artofproblemsolving.com/e/9/b/e9b98ba1a14efdf1cd888a6ae5f0d2de102e6d88.png)
Video Solution 1
~Education, the Study of Everything
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: ![\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} =\frac{\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7}}{\sqrt{\frac{8}{9}\cdot\frac{24}{25}\cdot\frac{48}{49}}} =\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} =\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{\textbf{(B)}\ 2}.\]](http://latex.artofproblemsolving.com/9/b/c/9bc609e60a9cfcb9d50c370e13198b7c9ae3a088.png)
~not_slay
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. 
