Difference between revisions of "2022 AMC 8 Problems/Problem 17"

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==Problem==
 
==Problem==
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If <math>n</math> is an even positive integer, the <math>\emph{double factorial}</math> notation <math>n!!</math> represents the product of all the even integers from <math>2</math> to <math>n</math>. For example, <math>8!! = 2 \cdot 4 \cdot 6 \cdot 8</math>. What is the units digit of the following sum? <cmath>2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!</cmath>
  
If <math>n</math> is an even positive integer, the <i>double factorial</i> notation <math>n!!</math> represents the product of all the even integers from <math>2</math> to <math>n</math>. For example, <math>8!! = 2 \cdot 4 \cdot 6 \cdot 8</math>. What is the units digit of the following sum? <cmath>2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!</cmath>
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<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8</math>
  
<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8\qquad</math>
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==Solution==
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Notice that once <math>n>8,</math> the units digit of <math>n!!</math> will be <math>0</math> because there will be a factor of <math>10.</math> Thus, we only need to calculate the units digit of <cmath>2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.</cmath> We only care about units digits, so we have <math>2+8+8+8\cdot8,</math> which has the same units digit as <math>2+8+8+4.</math> The answer is <math>\boxed{\textbf{(B) } 2}.</math>
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~wamofan
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/oUEa7AjMF2A?si=f4lLO32DQ4Yxdpkv&t=2925
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~Math-X
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==Video Solution (🚀Under 2 min🚀)==
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https://youtu.be/qOBzhBx8uw4
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 +
~Education, the Study of Everything
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 +
== Video Solution==
 +
https://youtu.be/wp9tOyJ3YQY?t=146
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 +
==Video Solution==
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https://youtu.be/Ij9pAy6tQSg?t=1461
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 +
~Interstigation
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 +
https://www.youtube.com/watch?v=FTVLuv_n9bY
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~Ismail.Maths
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==Video Solution==
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https://youtu.be/hs6y4PWnoWg?t=80
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~STEMbreezy
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==Video Solution==
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https://youtu.be/BbGqQaqE2rM
  
==Solution==
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~savannahsolver
  
Notice that once <math>n>8,</math> <math>n!!</math>’s units digit will be <math>0</math> because there will be a factor of <math>10</math>. Thus, we only need to calculate the units digit of <math>2!!+4!!+6!!+8!! = 2+2\cdot 4+2\cdot 4\cdot 6 + 2\cdot 4\cdot 6\cdot 8 = 2+8+48+48\cdot 8.</math> We only care about units digits so we have <math>2+8+8+8\cdot 8</math> Which is <math>2+8+8+4= \boxed{\textbf{(B) }2}</math>
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==Video Solution 8==
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https://www.youtube.com/watch?v=EVYrVkkpCo8
  
~wamofan
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~Jamesmath
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=16|num-a=18}}
 
{{AMC8 box|year=2022|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:04, 18 January 2024

Problem

If $n$ is an even positive integer, the $\emph{double factorial}$ notation $n!!$ represents the product of all the even integers from $2$ to $n$. For example, $8!! = 2 \cdot 4 \cdot 6 \cdot 8$. What is the units digit of the following sum? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]

$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$

Solution

Notice that once $n>8,$ the units digit of $n!!$ will be $0$ because there will be a factor of $10.$ Thus, we only need to calculate the units digit of \[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\] We only care about units digits, so we have $2+8+8+8\cdot8,$ which has the same units digit as $2+8+8+4.$ The answer is $\boxed{\textbf{(B) } 2}.$

~wamofan

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

https://youtu.be/oUEa7AjMF2A?si=f4lLO32DQ4Yxdpkv&t=2925

~Math-X

Video Solution (🚀Under 2 min🚀)

https://youtu.be/qOBzhBx8uw4

~Education, the Study of Everything

Video Solution

https://youtu.be/wp9tOyJ3YQY?t=146

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1461

~Interstigation

https://www.youtube.com/watch?v=FTVLuv_n9bY

~Ismail.Maths

Video Solution

https://youtu.be/hs6y4PWnoWg?t=80

~STEMbreezy

Video Solution

https://youtu.be/BbGqQaqE2rM

~savannahsolver

Video Solution 8

https://www.youtube.com/watch?v=EVYrVkkpCo8

~Jamesmath

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

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