Given the last $6$ digits of $M!-N!$, find $M(M-N)$ 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.
 A: 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.)
A: $$M!-N!=N!(\frac{M!}{N!}-1)$$
As the last three digits are zero, we conclude that $20>N>14$, because $(\frac{M!}{N!}-1)$ is not divisible by $10$. (not always but we assume that $M>>N$, so $\frac{M!}{N!}$ is divisible by 10)
Now you just need to put all the possible answers in $M(M-N)$ and try to find any natural M when $20>N>14$.
I didn't try all the variants but A cant be the answer because equation $M^2-MN-150=0$ has no natural solutions with the restrictions on N.
The possible answer is B.
