MHB What Are the Last Two Digits of the Sum of Factorials from 7! to 2016!?

[anemone]

Here is this week's POTW:

-----

What are the last two digits in $7! + 8! + 9! + ... + 2016 !$?

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. topsquark
2. kaliprasad
3. lfdahl

Solution from topsquark:

Spoiler

Since 10!, ... , 2016! are all multiples of 100 all we need to worry about are 7!, 8!, and 9! for the last two digits. Since 7! + 8! + 9! = 7! (1 + 8 + 8*9) = 5040*81 we have [math]5040 \cdot 81 \equiv 40 \cdot 81 \equiv 3240 \equiv 40 \text{ mod(100)}[/math] the last two digits must be 40.

(Or you could just calculate 7! + 8! + 9! = 408240.)