Difference between revisions of "2022 AMC 8 Problems/Problem 8"

Revision as of 23:37, 18 February 2022

Problem

What is the value of $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?$

$\textbf{(A) } \frac{1}{462} \qquad \textbf{(B) } \frac{1}{231} \qquad \textbf{(C) } \frac{1}{132} \qquad \textbf{(D) } \frac{2}{213} \qquad \textbf{(E) } \frac{1}{22}$

Solution 1

Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes $\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.$

~MRENTHUSIASM

Solution 2

The original expression becomes $\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.$

~hh99754539

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=565

~Interstigation

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.

@@ Line 15: / Line 15: @@
 ~hh99754539
+==Video Solution==
+https://youtu.be/Ij9pAy6tQSg?t=565
+~Interstigation
 ==See Also==
 {{AMC8 box|year=2022|num-b=7|num-a=9}}
 {{MAA Notice}}

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