 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Never 0 mod 3 (Posted on 2016-03-17) Let k be a positive integer. Suppose that the integers 1, 2, 3, ...3k, 3k + 1 are written down in random order.

What is the probability that at no time during this process, the sum of the integers that have been written up to that time is divisible by 3?

Source: Putnam competition

 No Solution Yet Submitted by Ady TZIDON No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Values for the first 5 k | Comment 2 of 7 | The following program counts instances directly by permuting strings of the form "1231231231", with k repetitions of "123" followed by a "1". All permutations are counted.

DefDbl A-Z
Dim crlf\$

Form1.Visible = True

Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)

For k = 1 To 5

s\$ = ""
For i = 1 To k
s = s + "123"
Next
s = s + "1"
h\$ = s
tr = 0: hit = 0
Do
tot = 0
tr = tr + 1
good = 1
For i = 1 To Len(s)
tot = tot + Val(Mid(s, i, 1))
If tot Mod 3 = 0 Then good = 0: Exit For
Next
If good Then hit = hit + 1
permute s
DoEvents
Loop Until s = h
Text1.Text = Text1.Text & k & Str(Len(s)) & Str(hit) & "/" & tr & mform(hit / tr, "  0.0000000") & crlf

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

End Sub

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

For the first five k, given below are k, the length of the string, and the probability in fraction and decimal:

1 4 3/12  0.2500000
2 7 15/210  0.0714286
3 10 84/4200  0.0200000
4 13 495/90090  0.0054945
5 16 3003/2018016  0.0014881

 Posted by Charlie on 2016-03-17 14:39:24 Please log in:

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