What is the remainder of $1!+2!+...+ (10^{10})! \mod40$?

Question

I want to know if my thinking is correct about this question!

The question: $1!+2!+...+ (10^{10})! \mod40$?

How \begin{align} 1 \equiv 1 \mod40\\ 2 \equiv 2 \mod40\\ 3 \equiv 3 \mod40\\ .\\ .\\ .\\ (40-1)\equiv (40-1) \mod40 \end{align} Then, If we multiplying term by term we have that: $3! \equiv 3! \mod40$, for example. On this, I can write $(10^{10})! \equiv (10^{10})! \mod40$.

So, $1!+2!+...+ (10^{10})! \equiv 1!+2!+...+ (10^{10})! \mod40$ and my answer is that remainder is: $1!+2!+...+ (10^{10})!$

Will be correct or is other thinking?

Answer 1

Now after all help, I get it! So, the answer is:

$1! + 2! +3!+ 4!+ 5!+...+ (10^{10})! \mod40$, how for the numbers bigger than 5 are multiples of 40, we have that $5!+...+ (10^{10})! \equiv 0\mod40$ and

\begin{align} 1! \equiv 1 \mod 40\\ 2! \equiv 2 \mod 40\\ 3! \equiv 6 \mod 40\\ 4! \equiv 24 \mod 40 \end{align}

then, $1!+2!+3!+4! \equiv33 \mod40$ and the remainder is: 33.

Answer 2

Hint: $n \geq 5$ has $n! \equiv 0 \pmod{40}$.