DefDbl A-Z

Dim crlf$

Function mform$(x, t$)

  a$ = Format$(x, t$)

  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$

  mform$ = a$

End Function

Private Sub Form_Load()

 ChDir "C:\Program Files (x86)\DevStudio\VB\projects\flooble"

 Text1.Text = ""

 crlf$ = Chr(13) + Chr(10)

 Form1.Visible = True

 DoEvents

 

 For n = 10000 To 99999

  s$ = LTrim(Str(n))

  prod = Val(Mid(s$, 2, 1)) * Val(Mid(s$, 4, 1))

  Sum = Val(Mid(s$, 1, 1)) + Val(Mid(s$, 3, 1)) + Val(Mid(s$, 5, 1))

  ct = ct + 1

  If Sum = prod Then

    goodct = goodct + 1

    Text1.Text = Text1.Text & n & Str(goodct) & Str(ct) & mform(goodct / ct, "#0.0000000") & crlf

    DoEvents

  End If

 Next

 Text1.Text = Text1.Text & Str(goodct) & Str(ct) & mform(goodct / ct, "#0.0000000") & crlf

 

 Text1.Text = Text1.Text & crlf & " done"

End Sub

finds

 1738 90000 0.0193111

 

which is 1738 such numbers out of the 90000 numbers in the range for a probability of 1738/90000 = 869/45000 = 0.019311111....

An analytic method would examine the 9*8/2=36 possible products of different digits and the 9 possible products of equal digits that could result in a positive sum, and see how many ways three single-digit numbers, the first of which can't be zero, can add up to the respective products. Any products greater than 27 could be ignored of course.

Needing consideration are:

1*1

1*2

1*3

1*4

1*5

1*6

1*7

1*8

1*9

2*2

2*3

2*4

2*5

2*6

2*7

2*8

2*9

3*3

3*4

3*5

3*6

3*7

3*8

3*9

4*4

4*5

4*6

5*5

Sets of digits adding to the products of the same integer (i.e., squares) count only once, while those of unequal digits count twice, Completion is "left as an exercise for the reader".