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

m (→‎Problem)
m (→‎Solution: Fixed errors like 3+8=1. I know that we care about the units digit only, but equations like this are incorrect.)
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==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>
+
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  
+
~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 23:15, 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? \[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]

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

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

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
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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