Large factorial sum
For any positive integer n, denote n! to be a factorial function in which it is the product of consecutive villagers from 1 to n. Find the last 2 digits of
0! + 1! + 2! + 3! + ... + (2^77232917 -1)! ?
Fun fact: 2^77232817 - 1 is the largest known Mersenne prime that was recently found last January 2018 by GIMPS or the Great Internet Mersenne Prime Search. It contains over 23 million digits!
The answer is 14.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
At first it can seems tricky, but it isn't.
The factorials that have 00 as leading zeroes is those that have 2 2 ⋅ 5 2 as one of it's factor. The first factorial with this property is 1 0 ! . After this point all factorials will have at least two leading zeroes. It means that the tens and unit digits will be 0.
Then we only need to sum up:
0 ! = 1
1 ! = 1
2 ! = 2
3 ! = 6
4 ! = 2 4
5 ! = 1 2 0
6 ! = 7 2 0
7 ! = 5 0 4 0
8 ! = 4 0 3 2 0
9 ! = 3 6 2 8 8 0
1 0 ! = 3 6 2 8 8 0 0 (not necessarily at all, I only included to show the 2 leading zeroes)
And the result will be 4 0 3 7 9 1 4 . If we keep going this sum until that huge factorial, the last two digits will always be 1 4 .