What is the value of $2019! \pmod{7}$?
I guess it's $0$? Because $$2019! = 2019\cdot2018\cdot2017\cdot ...\cdot7\cdot6\cdot...\cdot1$$ There's $7$ and also numbers that has $0$ remainder when divided by $7$, times the other numbers equals $0$.
Can someone correct me if I'm wrong? I'm still not sure this is the answer though. Thanks
Yes, $$2019! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times ... \times 2019$$ $$ =7 \times (1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 8 \times 9 \times ... \times 2019), $$ so $2019!$ is a multiple of $7$ and the remainder when $2019!$ is divided by $7$ is $0,$ and therefore we write $$2019! \equiv 0 \pmod 7.$$
Note that we do not need to compute $2019!$ -- a number with thousands of digits -- to make this determination.
As pointed out in the comments, indeed all natural numbers up to and including $2019$ divide $2019!,$ by a similar argument. Also, $7$ divides $2019!$ many times (exactly how many is a different question) because of the factors $14, 21, 28, ...$ in $2019!.$