Remainders Are Fun

Note that $\begin{aligned} \frac{1007 \times 1008 \times 1009 \times 1010 \times \cdots \times 2019}{1010!} &= 1007 \times 1008 \times 1009 \times \frac{1010\times 1011 \times \cdots \times 2019}{1010!} \\ &= 1007 \times 1008 \times 1009 \times \frac{2019!}{1010! \times 1009!} \\ &= 1007 \times 1008 \times 1009 \times \binom{2019}{1010} \in \mathbb{N} \end{aligned}$ so that $1010!$ evenly divides the product and $1007 \times 1008 \times 1009 \times 1010 \times \cdots \times 2019 \equiv \boxed{0} \pmod{1010!}.$

$\begin{aligned} \cfrac{1007\times1008\times1009\times \cdots \times 2019}{1013!} & = \cfrac{(1\times2\times \cdots \times1006)(1007\times1008\times \cdots \times2019)}{(1\times2\times \cdots \times1006)(1\times2\times \cdots \times1013)} \\ & = \cfrac{2019!}{(1006!)(1013!)} \end{aligned}$ $\begin{aligned} \cfrac{1007\times1008\times1009\times \cdots \times 2019}{1010!} & = \cfrac{(1\times2\times \cdots \times1006)(1007\times1008\times \cdots \times2019)}{(1\times2\times \cdots \times1006)(1\times2\times \cdots \times1013)} \times (1013\times 1012\times 1011) \\ & = \cfrac{2019! \times (1013\times 1012\times 1011)}{(1006!)(1013!)} \end{aligned}$

it means that: $\large \begin{aligned} ( 1007 \times 1008 \times 1009 \times 1010 \times \cdots \times 2019 ) \mod 1010! & \equiv \big( 2019! \times (1013\times 1012\times 1011) \big) \mod (1006! \times 1013!) \\ & \equiv \Big[ \left( (2019!) \mod ( 1006! \times 1013! )\right) \times \left( (1013\times 1012\times 1011) \mod ( 1006! \times 1013! ) \right)\Big] \mod ( 1006! \times 1013! ) \\ & \equiv (0 \times1) \mod (1006! \times 1013!) \\ & = \boxed{0} \end{aligned}$

The product of $r$ consecutive integers is divisible by $r!$ .

$\underbrace{1010\times{1011}\times{1012}\times \cdot \cdot \cdot \times{2019}}_{\text{1010 consecutive numbers}}$ is divisible by $1010!$ .

$\therefore$ Our answer is $\boxed{0}$

$\underline{NOTE}$

The terms $1007\times{1008}\times{1009}$ were extraneous.

So , we can even prove the number above to be divisible by $1013!$ .

$\underbrace{1007 \times 1008 \times 1009 \times 1010 \times \cdots \times 2019}_{\text{1013 consecutive numbers}}$ is divisible by $1013!$