Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
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+ | Note that common factors of the numerator and the denominator cancel, so 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> | ||
==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 10:57, 28 January 2022
Problem
What is the value of
Solution
Note that common factors of the numerator and the denominator cancel, so the original expression becomes
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.