The digits of either kind of pandigital number add up to 45.

45 is divisible by 9, and therefore all pandigital numbers are divisible by both 3 and 9. Some pandigital numbers are divisible by 2, and some of those by 4 and some of those are divisible by 8 (such as those ending in ...128). Those that are divisible by 2 are also divisible by 6 as all are divisible by 3. Any ending in 5 or 0 is divisible by 5.

7 is the first number attempted by the program.

It turns out 100, 200 and 300 are the first three, as each requires the number end with two zeros.

Likewise, when using only digits 1 through 9, all the multiples of 10 fail, including 10 as the first.

Private Sub Form_Load()

 Text1.Text = ""

 crlf$ = Chr(13) + Chr(10)

 Form1.Visible = True

 

 s$ = "123456789"

 h$ = s

 Do

   v = Val(s)

   For i = 7 To 300

     q = Int(v / i): r = v - q * i

     If r = 0 Then

       If dvsr(i) = 0 Then ct = ct + 1

       dvsr(i) = 1

     End If

   Next i

   permute s

   DoEvents

 Loop Until s = h Or ct = 294

 

 If ct = 294 Then

   Text1.Text = Text1.Text & "not found" & v

 Else

   For i = 7 To 300

     If dvsr(i) = 0 Then Text1.Text = Text1.Text & Str(i)

   Next

   Text1.Text = Text1.Text & crlf

 End If