mathematics - How many digits can be removed from a division puzzle?

Similarly to How many digits can be removed from a multiplication puzzle and still give only one answer? I am curious about division puzzles.

I know this puzzle:

.

It has only 2 out of 42 digits $\approx$ 4.8% digits shown.

Is there a puzzle with less percent of digits shown?

Answer

You are looking for least percentage, without concerning with the Difficulty of the puzzle.

In that case, the percentage can be made arbitrarily small. Here is an example with 31 digits removed:

999)*********(*******
    ***
       ***

       ***
          ***
          ***
            0

Here the format AAA)BBB(CCC means AAA is dividing BBB to get the answer as CCC.

We can extend the same arbitrarily, eg here is the next possibility with 43 digits removed:

999)************(**********
    ***

       ***
       ***
          ***
          ***
             ***
             ***
               0

We now list the third example with 55 digits removed:

999)***************(*************
    ***
       ***
       ***
          ***
          ***
             ***
             ***
                ***
                ***

                  0

We now list a fourth example with 67 digits removed:

999)******************(****************
    ***
       ***
       ***
          ***
          ***

             ***
             ***
                ***
                ***
                   ***
                   ***
                     0

Enough Examples, let us generalise this, such that 999 Divides a (3N)-Digit number with a (3N-2)-Digit number as the answer.

Calculating the percentage:

999)************[3N '*' characters](**********[3N-2 '*' characters]

This will contain totally (3N)+(3N-2)+(3N)+(3N-3) = 12N-5 '*' characters, when counting all the rows.

With 4 known digits [9,9,9,0] , total number of digits = 12N-1.

Percentage of known digits = 100*(4)/(12N-1).

We can take N = 100, for example, to get 0.3%.

We can take N = 1000 to get 0.03%.

EDIT:

Solutions to those puzzles will be like this: The result Digits will be 1001001...001 (1 followed by 001 (N-1) times). All remaining unknown Digits will be 9s.

999)999999999(1001001
    999
       999
       999
          999
          999
            0

999)999999999999(1001001001
    999
       999
       999
          999
          999
             999
             999
               0