Face up the problem bravely

Easy too check that $2017$ satisfies properties (1) and (2). And I'll prove it's the largest in the following discussion.

Proof by contradiction.

1. Assume $2017$ is not the largest.

Then there exists a smallest integer that larger than $2017$ and satisfies properties (1) and (2),say, $c$ ,

i.e._(1): $c$ is a divisor of $(2018!)!$ and _(2):any element in set $A= \{ x∈Z|2018≤x≤2018! \}$ is not a divisor of $c$ .

2. We see that $c$ is larger than $2018!$ . Because, if not, $c$ will be someone in set $A$ , thus $c$ is not equipped with property (2),contradiction.

So since $c≥2018!+1$ and $c$ divides $(2018!)!$ , $c$ cannot be a prime number , as every prime factor of $(2018!)!$ is no larger than $2018!$ .

3. Suppose one prime factor of $c$ is $p$ .

Because any number in $A$ ,say $k$ , is not a divisor of $c$ , $k$ cannot divide $p$ and $\frac{c}{p}$ .

Besides,since $c$ divides $(2018!)!$ , p and $\frac{c}{p}$ divide $(2018!)!$ ,too.

Thus p and $\frac{c}{p}$ have properties (1) and (2).

4. Now,since c is the smallest integer that larger than $2017$ and satisfies (1) and (2),

and note $p$ and $\frac{c}{p}$ are less than $c$ , $p$ and $\frac{c}{p}$ should both less than or equal to 2017 ,

which leads to $c={p}×\frac{c}{p}≤2017^{2}$ . But remember $c$ $>2018!>2017^{2}$ . Contradiction! Thus the $\boxed{2017}$ is the largest. $\Box$

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General Situation .

Given $m$ as an integer larger than $2$ and $n$ as an integer larger than $m(m-2)$ .Determine the largest integer $d$ satisfying following properties:

(1) $d|n!$ ,

(2)any element in set $A= \{ x∈Z|m≤x≤n\}$ is not a divisor of $d$ .

——-——Similarly,the answer is $m-1$ .