5
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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?

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  • 1
    $\begingroup$ I have seen a similar puzzle without the last zero. That reduces the count yet farther. $\endgroup$ – Ross Millikan Jun 15 '14 at 22:43
8
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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
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  • 1
    $\begingroup$ What do ")" and "(" mean? $\endgroup$ – klm123 Apr 4 '15 at 9:30
  • $\begingroup$ That is one way to write the Divisions: "999)999999(1001" means 999 is Dividing 999999, and the answer is 1001. $\endgroup$ – Prem Apr 4 '15 at 9:32
  • $\begingroup$ In your representation, you would have written 999999 on the left, 999 on the right, and 1001 below that. $\endgroup$ – Prem Apr 4 '15 at 9:36
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    $\begingroup$ Looks like someone's going for a Necromancer badge ;-) $\endgroup$ – Rand al'Thor Apr 4 '15 at 10:54
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    $\begingroup$ I see. Could you write few words about solution? $\endgroup$ – klm123 Apr 4 '15 at 14:08
1
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In More Mathematical Puzzles and Diversions, ch. 14, Nine More Problems, problem 5, Martin Gardner gives the problem of reconstructing the following long division:

       xx8xx
xxx)xxxxxxxx
     xxx
     ---
      xxxx
       xxx
      ----
        xxxx
        xxxx
        ----

Gardner credits the problem to P. L. Chessin of the Westinghouse Electric Corporation, and cites the April 1954 issue of American Mathematical Monthly. The answer is

10020316/124=80809

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