Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | Note that common factors (from <math>3</math> to <math>20,</math> inclusive) 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 | ||
Revision as of 14:58, 28 January 2022
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
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. 
