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

Question

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?

Answer 1

Hint: $6!$ is already divisible by $36$.

Answer 2

The idea is to find the smallest positive integer $x$ such that $36|(x!)$ as $(m!)| (n!)$ for $m\le n$

if $36|(x!), 36$ will divide $y!$ for all $y\ge x$

Here $m,n,x,y$ are assumed to be positive integers

Now, by inspection, $\displaystyle5!=120\equiv12\pmod{36},\implies 6!=6\cdot120\equiv6\cdot12\equiv0\pmod{36}$

Answer 3

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.