Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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<math>\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}</math> | <math>\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Note that common factors of the numerator and the denominator cancel, | + | Note that common factors (from <math>3</math> to <math>20,</math> inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes <cmath>\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.</cmath> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | The original expression becomes <cmath>\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | ||
| + | ~hh99754539 | ||
| + | |||
| + | ==Video Solution by Math-X (First understand the problem!!!)== | ||
| + | https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996 | ||
| + | |||
| + | ~Math-X | ||
| + | |||
| + | ==Video Solution (CREATIVE THINKING!!!)== | ||
| + | https://youtu.be/vr7hf7IK2po | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565 | ||
| + | |||
| + | ~Interstigation | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/1xspUFoKDnU?t=176 | ||
| + | |||
| + | ~STEMbreezy | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/Zgs4B6ajyOY | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/ITVgbv6MoVA | ||
| + | |||
| + | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=7|num-a=9}} | {{AMC8 box|year=2022|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:40, 23 November 2023
Contents
Problem
What is the value of ![\[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?\]](http://latex.artofproblemsolving.com/8/e/4/8e422ea624f4cc14204e47ec836017b740d90c3d.png)
Solution 1
Note that common factors (from 
to 
inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes ![\[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.\]](http://latex.artofproblemsolving.com/2/2/3/223017c5281b22f6c831d21b4c389a60b0d57a19.png)
~MRENTHUSIASM
Solution 2
The original expression becomes ![\[\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.\]](http://latex.artofproblemsolving.com/d/a/6/da6e9d851413902746f298d75123fb091283402b.png)
~hh99754539
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565
~Interstigation
Video Solution
https://youtu.be/1xspUFoKDnU?t=176
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
~harungurcan
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
