2022 AMC 8 Problems/Problem 17

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

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

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2022 AMC 8 Problems/Problem 17

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