2022 AMC 8 Problems/Problem 17

Revision as of 23:12, 28 January 2022 by MRENTHUSIASM (talk | contribs) (Problem)

Problem

If $n$ is an even positive integer, the 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 $n>8,$ $n!!$’s units digit will be $0$ because there will be a factor of $10$. Thus, we only need to calculate the units digit of $2!!+4!!+6!!+8!! = 2+2\cdot 4+2\cdot 4\cdot 6 + 2\cdot 4\cdot 6\cdot 8 = 2+8+48+48\cdot 8.$ We only care about units digits so we have $2+8+8+8\cdot 8$ Which is $2+8+8+4= \boxed{\textbf{(B) }2}$

~wamofan

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions

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