Number Theory

$\begin{aligned} 2017!+2018!+2019! & = 2017!(1+2018+2018 \times 2019) \\ & = 2017!(2019+2018 \times 2019) \\ & = 2017! (2019) (1+2018) \\ & = 2017! (2019^2) \end{aligned}$

Since 2019 is divisible by 3 (its sum of digits is divisible by 3), it is not a prime. But $\boxed{2017}$ is a prime and it is the largest prime factor.

The above expression can be factorized as $2017!(1+2018+2018 \times 2019) = 2017! \times 2 \times 2019$ . Now The $1+ 2018+ 2018 \times 2019 = 4076361$ Now $40766361= 3^{2} \times 673^{2}$ and $2017 > 673$

And $2017$ is a prime $\therefore$ Answer comes out to be $2017$