2017-digit multiple of 7!?

The largest is

$a=7.\lfloor \frac{10^{2017}}{7} \rfloor$

note that

$10^{2017}=7\times t+3 \implies t=\frac{10^{2017}-3}{7} \implies a=7 \times t=10^{2017}-3$

$a$ has 2016 $9$ 's and 1 $7$ , so the sum of its digits is $A=9.2016+7$

Similarly, the smallest is

$b=7.(\lfloor \frac{10^{2016}}{7}+1) \rfloor$

and

$10^{2016}=7\times k+1 \implies k=\frac{10^{2016}-1}{7} \implies b=7 \times (k+1)=10^{2016}-1+7=10^{2016}+6$

$b$ has 2015 $0$ 's, 1 $1$ and 1 $6$ , so the sum of its digits is $B=1+6$

Finally $A-B=2016 \times 9$

2017=(6)(336)+1. Each 6n digit multiple of 7 (n is a positive integer) contains 6n 9's, and (6n+1) digit multiple contains 6n 9's and a 7 in the end. Therefore A=(2016)(9)+7=18151. The minimum 2017 digit multiple is 7 more than the maximum 2016 digit multiple of 7 by 7. Therefore this minimum has 1 as the leftmost digit followed by 2015 zeros and 6 as the unit place digit. Therefore B=1+6=7 and hence A-B=18151-7=18144