Difference between revisions of "2022 AMC 10B Problems/Problem 5"
Mathboy100 (talk | contribs) (→Solution 2.1 (Brute Force + Cancellation)) |
MRENTHUSIASM (talk | contribs) (→Solution 2.1 (Brute Force + Cancellation)) |
||
| Line 22: | Line 22: | ||
~not_slay | ~not_slay | ||
| − | ==Solution | + | ==Solution 3 (Brute Force)== |
This solution starts off exactly as the one above. We simplify to get: | This solution starts off exactly as the one above. We simplify to get: | ||
| Line 29: | Line 29: | ||
But now, we can get a nice simplification as shown: | But now, we can get a nice simplification as shown: | ||
| − | |||
| − | |||
<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)}}} | <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}} | ||
=\dfrac{4\cdot6\cdot8}{3\cdot5\cdot7} \hspace{0.05 in} \cdot \hspace{0.05 in}\dfrac{3\cdot5\cdot 7}{2^5 \cdot 3} | =\dfrac{4\cdot6\cdot8}{3\cdot5\cdot7} \hspace{0.05 in} \cdot \hspace{0.05 in}\dfrac{3\cdot5\cdot 7}{2^5 \cdot 3} | ||
| − | =\dfrac{2^6\cdot 3}{2^5\cdot 3} = \boxed{\ | + | =\dfrac{2^6\cdot 3}{2^5\cdot 3} = \boxed{\textbf{(B)}\ 2}.</cmath> |
~TaeKim | ~TaeKim | ||
Revision as of 12:11, 28 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)
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
Solution 3 (Brute Force)
This solution starts off exactly as the one above. We simplify to get:
![\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}}\]](http://latex.artofproblemsolving.com/2/7/a/27a39e17394ca808a8f128d3a64376ff4f6d7e06.png)
But now, we can get a nice simplification as shown:
![\[\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{2^{10} \cdot 3^{2}}{3^2\cdot 5^2\cdot 7^2}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\frac{2^5 \cdot 3}{3\cdot5\cdot 7}} =\dfrac{4\cdot6\cdot8}{3\cdot5\cdot7} \hspace{0.05 in} \cdot \hspace{0.05 in}\dfrac{3\cdot5\cdot 7}{2^5 \cdot 3} =\dfrac{2^6\cdot 3}{2^5\cdot 3} = \boxed{\textbf{(B)}\ 2}.\]](http://latex.artofproblemsolving.com/0/b/9/0b90fdafc0f990819d834b40155c5a312336f31f.png)
~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. 
