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Since we are dividing by a single digit integer, our maximum remainder can be $8$

So let us analyse whether we can achieve remainder of $8$ or not

Case 1: Remainder of $8$

We have to divide by $9$ to achieve a remainder = $8$

The sequence of numbers $17,26,35,44,53,62,71,80,89,98$ produces remainder = $8$ when divided by $9$

We notice that in this sequence upto $80$ , if $\overline{ab}$ is divided by $9$ then we need to multiply $9$ by $a$ which is a repetition of digit $a$ and not allowed.

Now $89$ and $98$ both contain $8$ in them and reimainder is also $8$ but this contradicts that all digits written in long division must be distinct.

So a remainder of 8 is not possible .

Case 2: Remainder of $7$

First we try to divide by $9$

The sequence of numbers $16,25,34,43,52,61,70,79,88,97$ produces a remainder of $7$ on division by $9$

We remove $70,79,88,97$ from our list as $7$ cannot both be in the number and the remainder and in $88$ we have already $8$ repeated two times.

Upto $61$ by the same previous logic we need to multiply $9$ by the first digit of that number, example, $61=9 \times 6+7$ Here $6$ is repreated and hence is not accordance with the rules

Now we check that can we get remainder of $7$ when we divide by $8$

Here the required sequence is $15,23,31,39,47,55,63,71,79,87,95$

we remove $47,71,79,87$ because they contain $7$ and remainder is also $7$ .This repeats $7$ and is not allowed. $55$ is also removed as it has $5$ two times.

$8 \times 7+7=63$ is also removed because we divide by $7$ and remainder is also $7$ which is not allowed.

We are left with $15,23,31,39,95$

Now for $15,23,31$ we need to multiply $8$ by their first digits which forces a repetition of $1,2,3$ respectively. For $39=8 \times 4+7= 32+7$ in which $3$ is repeated. In case of $95$ we need $8$ to multiply by $11$ so that $1$ is repeated.

In this case also we cannot get a remainder of $7$

Conclusion: Remainder of $7$ is also not possible to achieve

Case 3: Remainder of 6

Remainder of 6 is achieved in this way

$90=7 \times 12 +6 = 84+6$

So the answer is $\boxed{6}$ since we checked that no one of the greater remainders are possible.

NOTE: If I can write long division box like @Pi Han Goh then my solution will be better but I don'tknow how to write. Can anybody suggest how to make this long division box and write numbers in it.