Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
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| + | ==Solution 2== | ||
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| + | <math>\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22} = \frac{20!}{\frac{22!}{2}}</math> = \frac{20! \cdot 2}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \frac{1}{231} \implies \boxed{\textbf{(B) } \frac{1}{231}}$ | ||
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| + | ~hh99754539 | ||
==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}} | ||
Revision as of 18:32, 31 January 2022
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
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
= \frac{20! \cdot 2}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \frac{1}{231} \implies \boxed{\textbf{(B) } \frac{1}{231}}$
~hh99754539
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. 
