Remainder of the sum of consecutive factorials

Question

$1! + 2! + 3! + \cdots + 2019!$

What is the remainder when divided by $15$?

I don't know how to work out this problem. Is there any shortcut to find the remainder of this kind of problems?

Answer 1

$$ n! = n(n-1)(n-2)(n-3)\cdots\cdots6\cdot{} \underbrace{5\cdot4\cdot3\cdot2\cdot1} $$ The expression over the $\underbrace{\text{underbrace}}$ is a multiple of $15.$

So $n!$ is always a multiple of $15$ except $4!=4\cdot3\cdot2\cdot1,\quad$ $3!= 3\cdot2\cdot1,\quad$ $2!=2\cdot1, \quad$ $1! = 1, \quad$ $0!=1.$ So you can ignore the sum of those larger factorials.

Answer 2

The biggest shortcut is that if $a < b$ and $m\ge b$ then $m! = 1\times2\times\cdots\times a \times \cdots \times b\times\cdots \times m$ is divisible by $ab$.

So for any $m \ge 5$ we have $m! =1\times 2 \times \color{blue} 3 \times 4 \times\color{blue}{5} \times \cdots \times m$ so $15\mid m!$ for all $m\ge 5$.

So $15\mid 5! + 6! + \cdots + 2019!$ so the remainder of $1+2!+3!+4!+5! + \cdots +2019!$ is the same as the remainder of $1!+2! +3! +4!$ which is .... easy to figure out.