... restate the problem as abcd = efgh * i

this can go a bit further:

abcd = efgh * i

a * 10^3 + b * 10^2 + c * 10^1 + d * 10^0 =
      (e * 10^3 + f * 10^2 + g * 10^1 + h * 10^0) * i

=> (e*i-a)*10^3 + (f*i-b)*10^2 + (g*i-c)*10^1 + (h*i-d)*10^0 = 0

for the above to have a chance to be zero:
1)
  ((h*i-d) * 10^0) % 10^1 = 0  (e.g. must end in 0)
2)
  The cumulative sum of the above at position J,
  (i.e. from right to left, J=0 for the term (h*i-d) * 10^0 )
  must also end in zeros as follows:
   cum-sum at pos J=1 must end in 00
   cum-sum at pos J=2 must end in 000
    etc.
3)
  The cumulative sum of the above at position J,
  from right to left must also not be less than:
    cum-sum at pos J+1 >= 100
    cum-sum at pos J+2 >= 1000
      etc.

[download]

As an example, for the fraction: 6952 / 1738 = 4 it is true that:

   (8*4-2)*10^0=30
              ...     (ends in 0),

   ((8*4-2)*10^0+(3*4-5)*10^1)=100
              ...     (ends in 00 and is not less than 10^2)

   ((8*4-2)*10^0+(3*4-5)*10^1+(7*4-9)*10^2)=2000
              ...     (ends in 00 and is not less than 10^3)

     etc.

[download]

For base-10 the first rule skips 90% of the cases. And each subsequent rule 90% of the remaining. E.g. 362880 total cases -> 25920 -> 2880 (I used permutations and took the pivot as the missing digit. That can perhaps get better).

In reply to Re: Efficient enumeration of pandigital fractions by bliako
in thread Efficient enumeration of pandigital fractions by kikuchiyo

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