To find: Last digit of 1! + 2!+ .... + n!

IMP:
1! = 1
1! + 2! = 1 + 2 = 3
1! + 2! + 3! = 1 + 2 + 6 = 9
1! + 2! + 3! + 4! = 9 + 24 = 33
1! + 2! + 3! + 4! + 5! = 33 + 120 = _ _ 3
Now onwards.. Unit digit will remain "3" only as each factorial after 5! will be "0" as unit digit.
Example: 6! = 720

Statement 1:
n is divisible by 5 ==> n = 5, 10
Unit Digit = 3

Statement 2:
(n+1)/6
Minimum possible value of n = 5
Unit Digit = 3

Answer [spoiler]{D}[/spoiler]?

psm12se wrote:If n is a positive integer, what is the last digit of 1! + 2!+ .... + n!?

1. n is divisible by 5

2. (n+1)/6 is a positive integer.

I solved this problem but took 10 mins to solve it. Is there a way to solve it in 2 mins. I used the plugin method as a approach.

Hi Psm12se,

Last Digit of 1!+2!+3!+........n!
Always take 3 values
When N=1 last digit = 1
When N=3 last digit = 9
for the rest of values last digit will be 3.

Statement 1 : n is divisible by 5
So from above we can conclude last digit will always be 3.
Sufficient.

Statement 2: (n+1)/6 is a positive integer.
take any positive integer like 1 or 2 or anything.
you will end up with last digit as 3.
Sufficient.

Hence Answer D

Regards,
Uva.