TITLE: Given the last $6$ digits of $M!-N!$, find $M(M-N)$
QUESTION [6 upvotes]: If the last 6 digits of $M!-N!$ are $999000$, which of the following option is not possible for $M\times(M-N)$? 
A- 150
B- 180
C- 200
D- 225
E- 234
Both $M$ and $N$ are positive integers and $M>N$. $M!$ is factorial of $M$.
How to solve this question? I am unable to make any progress.

REPLY [2 votes]: I'm turning a comment into an answer.
If the last six digits of $M!-N!$ are $999000$ (with $M\gt N$), then $N!$ is divisible by $8$ but not $16$, which means $N=4$ or $5$.  But that implies $M!$ ends in either $999120$ or $999024$, neither of which is possible for a factorial.  (To end with that big a remainder, $M$ would have to be greater than $9$, but if $M\ge10$, then $M!$ ends with two $0$'s.)  So paradoxically, perhaps, we can conclude that all of the options for $M(M-N)$ are possible, because the premise is always false.
(Thanks to Michael for pointing out a flaw in the original answer.)