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

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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>
 
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>
  
<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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<math>\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8\qquad</math>
  
 
==Solution==
 
==Solution==
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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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~wamofan
  
 
==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}}

Revision as of 14:11, 28 January 2022

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? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]

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

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
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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