2022 AMC 8 Problems/Problem 8
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
The key observation in this problem is that all of the numbers from 
to 
(inclusive) cancel out. To see this, notice that for all numbers to (inclusive) there's one of them in a numerator and another in a denominator. Thus, the entire expression simplifies to
![\[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22} = \frac{1\cdot 2}{21 \cdot 22} = \frac{1}{21\cdot 11}.\]](http://latex.artofproblemsolving.com/8/b/6/8b64c534dd42aca862441bf740ca521fa33f764f.png)
Thus, our answer is 
Note that common factors of the numerator and the denominator cancel, so 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. 
