In number 2019! there's included 2019*2018*2017...*100*99*98...*10*9*8*7...*1 so when 10*100*1000=1000000 so there could be found 6 zeros in 2019!. When it is divided by 2^1009 which its last three digits are 912 means that 000/912=000 in the end the last three digits of 2019!/2^1009=000(mod 1000)
Firstly we have that 1000=8*125, because 2019! is divisible by 125 and if we prove that our expreesion is divisible by 8 we end up with the result that last 3 digits are 000.So if 2019 is divisble by 2^(1009+3) we are done.By Legendre formula we see that 2019! has more than 2000 2s.So it is clearly divisible by 2^1012.Last three digits are 000