7
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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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  • 2
    $\begingroup$ I have seen a similar puzzle without the last zero. That reduces the count yet farther. $\endgroup$ Commented Jun 15, 2014 at 22:43

3 Answers 3

9
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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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    $\begingroup$ What do ")" and "(" mean? $\endgroup$
    – klm123
    Commented Apr 4, 2015 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
    Commented Apr 4, 2015 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
    Commented Apr 4, 2015 at 9:36
  • 1
    $\begingroup$ Looks like someone's going for a Necromancer badge ;-) $\endgroup$ Commented Apr 4, 2015 at 10:54
  • 1
    $\begingroup$ I see. Could you write few words about solution? $\endgroup$
    – klm123
    Commented Apr 4, 2015 at 14:08
5
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Remarkably, there are long division problems (WITH NO DIGITS GIVEN!) that have a unique solution.

The following problem was given in the Eureka journal (April 1978/University of Cambridge). This problem can be solved by hand without computers.

Long division puzzle with all digits missing

Unique solution:

$\Large\frac{631938}{625}$

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1
  • $\begingroup$ Great puzzle but decimal point is a bit of a game changer though $\endgroup$
    – Laska
    Commented Aug 27, 2022 at 1:16
3
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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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