Finding Belgian numbers

[Eugene McDonnell]

If n is a positive integer                          (176)
and s is the sum-scan of its digits                 (1 8 14)
and m is the sum of its digits                      (14)

find j, the number of integral times m goes into n
  j = floor(n/m)                                     (12)

find k, the multiple of n closest below or at n
k = j * m                                           (168)

and p, the sum of k and s                           (169 176 182)

then, if n is in p, it is Belgian, and not otherwise.

Another example:
   n is 1234567898765
   s is 1 3 6 10 15 21 28 36 45 53 60 66 71
   m is 71
   j is 17388280264
   k is 1234567898744
   p is 1234567898745 1234567898747 1234567898750 1234567898754 
1234567898759 1234567898765 1234567898772 1234567898780 1234567898789 
1234567898797 1234567898804 1234567898810 1234567898815
   n is equal to the sixth item of p and so is Belgian.

Conversely, using the inverse of this logic, if (n - s)mod(m) contains 
a 0, n is Belgian.

    n - s
1234567898764 1234567898762 1234567898759 1234567898755 1234567898750 
1234567898744 1234567898737 1234567898729 1234567898720 1234567898712 
1234567898705 1234567898699 1234567898694

Notice that the sixth difference is k from the previous example.

   (n - s) mod m
20 18 15 11 6 0 64 56 47 39 32 26 21

The sixth item is 0, so n is Belgian.

Eugene McDonnell
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