Difference between revisions of "2022 AMC 10B Problems/Problem 5"
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<math>\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}</math> | <math>\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}</math> | ||
| − | ==Solution== | + | ==Solution 1 (Difference of Squares)== |
We apply the difference of squares to the denominator, and then regroup factors: | We apply the difference of squares to the denominator, and then regroup factors: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | \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)}} &= \frac{1+\frac13}{\sqrt{1+\frac13 | + | \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)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ |
| − | &= \ | + | &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ |
| + | &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ | ||
| + | &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ | ||
| + | &= \frac{\sqrt8}{\sqrt2} \\ | ||
| + | &= \boxed{\textbf{(B)}\ 2}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==Solution 2 (Brute Force)== | ||
| + | Since these numbers are fairly small, we can use brute force as follows: <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}{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}.</cmath> | ||
| + | ~not_slay | ||
| + | |||
| + | ==Solution 3 (Brute Force)== | ||
| + | |||
| + | 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> |
| + | |||
| + | 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)}}} | ||
| + | = \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}.</cmath> | ||
| + | |||
| + | ~TaeKim | ||
| + | |||
| + | ~minor edits by mathboy100 | ||
| + | |||
| + | ==Video Solution (⚡️2 min solution⚡️)== | ||
| + | https://youtu.be/N7hGuy0MWOQ | ||
| + | |||
| + | ~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 ![\[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\]](http://latex.artofproblemsolving.com/e/9/b/e9b98ba1a14efdf1cd888a6ae5f0d2de102e6d88.png)
Solution 1 (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: ![\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} =\frac{\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7}}{\sqrt{\frac{8}{9}\cdot\frac{24}{25}\cdot\frac{48}{49}}} =\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} =\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{\textbf{(B)}\ 2}.\]](http://latex.artofproblemsolving.com/9/b/c/9bc609e60a9cfcb9d50c370e13198b7c9ae3a088.png)
~not_slay
Solution 3 (Brute Force)
This solution starts precisely like the one above. We simplify to get the following:
![\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}}\]](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 (⚡️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. 
