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

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<math>\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}</math>
 
<math>\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}</math>
  
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==Solution==
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==Solution 1==
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The key observation in this problem is that all of the numbers from <math>3</math> to <math>20</math> (inclusive) cancel out. To see this, notice that for all numbers <math>3</math> to <math>20</math> (inclusive) there's one of them in a numerator and another in a denominator. Thus, the entire expression simplifies to 
 
−
<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\cdot 2}{21 \cdot 22} = \frac{1}{21\cdot 11}.</cmath>
 
  
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Thus, our answer is <math>\boxed{\textbf{(B) } \frac{1}{231}}</math>
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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>
  
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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>
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~MRENTHUSIASM
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 +
==Solution 2==
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The original expression becomes <cmath>\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}}.</cmath>
 +
~hh99754539
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/vr7hf7IK2po
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution==
 +
https://youtu.be/1xspUFoKDnU?t=176
 +
 
 +
~STEMbreezy
 +
 
 +
==Video Solution==
 +
https://youtu.be/Zgs4B6ajyOY
 +
 
 +
~savannahsolver
 +
 
 +
==Video Solution==
 +
https://youtu.be/ITVgbv6MoVA
 +
 
 +
~harungurcan
  
−
~MRENTHUSIASM
+
==Video Solution by Dr. David==
 +
https://youtu.be/4TWIyoKdOcU
  
 
==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}}

Latest revision as of 20:29, 25 November 2024

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 by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/vr7hf7IK2po

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565

~Interstigation

Video Solution

https://youtu.be/1xspUFoKDnU?t=176

~STEMbreezy

Video Solution

https://youtu.be/Zgs4B6ajyOY

~savannahsolver

Video Solution

https://youtu.be/ITVgbv6MoVA

~harungurcan

Video Solution by Dr. David

https://youtu.be/4TWIyoKdOcU

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. AMC logo.png

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