If the sum $\sum_{x=1}^{100}x!$ is divided by $36$, how to find the remainder? 
If the sum $$\sum_{x=1}^{100}x!$$ is divided by $36$, the remainder is $9$.

But how is it?
THIS said me that problem is $9\mod 36$, but how did we get it?

One should consider that numbers written in a base are a series of remainders, so for example, 1957 gives 195 remainder 7, and 195 gives 19 remainder 5, and 19 is 1 remainder 9.  So 1957 is 1 re 9 re 5 re 7.  Adding remainders is then like adding the last n places of a base.
For $n$!, we have all factorials 6! or larger, are multiples of 36, so it's just a matter of adding the first 5 (1+2+6+24+120 = 153), and dividing this by 36 (4 remainder 9).  It's only the 9 we want.  
