2022 AMC 8 Problems/Problem 17: Difference between revisions
No edit summary |
|||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | == Problem == | ||
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 <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> | ||
<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</math> | ||
==Solution== | == Solution 1 == | ||
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> | 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> | ||
| Line 10: | Line 11: | ||
~wamofan | ~wamofan | ||
== | ==Solution 2 (Solution 1 worded differently)== | ||
We can see that after <math>8!!</math> in the sequence, the units digit is always <math>0</math> (every value after <math>8!!</math> is a multiple of <math>10</math>). Therefore, our answer is the sum of the units digits of <math>2!!, 4!!, 6!!,</math> and <math>8!!</math> respectively. This sum is equal to <math>2 + 8 + 8 + 4</math>, or <math>\boxed{\textbf{(B) } 2}.</math> | |||
~Irfans123 | |||
~Education, the Study of Everything | == Video Solution 1 == | ||
[//youtu.be/tYWp6fcUAik?si=V8hv_zOn_zYOi9E5&t=2049 ~hsnacademy] | |||
== Video Solution 2 == | |||
[//youtu.be/oUEa7AjMF2A?si=f4lLO32DQ4Yxdpkv&t=2925 ~Math-X] | |||
==Video Solution 3 == | |||
[//youtu.be/qOBzhBx8uw4 ~Education, the Study of Everything] | |||
== Video Solution 4 == | |||
https://youtu.be/wp9tOyJ3YQY?t=146 | https://youtu.be/wp9tOyJ3YQY?t=146 | ||
==Video Solution== | == Video Solution 5 == | ||
[//youtu.be/Ij9pAy6tQSg?t=1461 ~Interstigation] | |||
== Video Solution 6 == | |||
[//youtu.be/hs6y4PWnoWg?t=80 ~STEMbreezy] | |||
~ | == Video Solution 7 == | ||
[//youtu.be/BbGqQaqE2rM ~savannahsolver] | |||
==Video Solution== | == Video Solution 8 == | ||
~ | [//www.youtube.com/watch?v=EVYrVkkpCo8 ~Jamesmath] | ||
==Video Solution== | == Video Solution 9 == | ||
https://youtu.be/1Vg8Mt0bSbQ | |||
==Video Solution | == Video Solution 10 == | ||
~ | [//youtube.com/FTVLuv_n9bY ~Ismail.Maths] | ||
== | == 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}} | ||
[[Category:Introductory Number Theory Problems]] | |||
Latest revision as of 01:14, 5 March 2025
Problem
If 
is an even positive integer, the 
notation 
represents the product of all the even integers from 
to . For example, 
. What is the units digit of the following sum? ![\[2!! + 4!! + 6!! + \cdots + 2018!! + 2020!! + 2022!!\]](http://latex.artofproblemsolving.com/6/b/7/6b7552c1c8d367a79bf701d4631efea748362631.png)
Solution 1
Notice that once 
the units digit of will be 
because there will be a factor of 
Thus, we only need to calculate the units digit of ![\[2!!+4!!+6!!+8!! = 2+8+48+48\cdot8.\]](http://latex.artofproblemsolving.com/7/e/2/7e269223baea7ee3fc6200b9d06bb91f0c38b27e.png)
We only care about units digits, so we have 
which has the same units digit as 
The answer is 
~wamofan
Solution 2 (Solution 1 worded differently)
We can see that after 
in the sequence, the units digit is always (every value after is a multiple of 
). Therefore, our answer is the sum of the units digits of 
and respectively. This sum is equal to 
, or
~Irfans123
Video Solution 1
Video Solution 2
Video Solution 3
~Education, the Study of Everything
Video Solution 4
https://youtu.be/wp9tOyJ3YQY?t=146
Video Solution 5
Video Solution 6
Video Solution 7
Video Solution 8
Video Solution 9
Video Solution 10
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
