A five digit positive integer N without a leading zero is selected at random.
Determine the probability that the sum of the first digit, third digit and fifth digit of N is equal the product of its second digit and fourth digit.
DefDbl AZ
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 singledigit 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".

Posted by Charlie
on 20140922 16:29:18 