All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Magic matrix (Posted on 2015-10-26)
The 1 to 9 digits are randomly arranged into a 3x3 array.

Find the probability that the sum of the numbers in every row, column, and diagonal is a multiple of 9.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 computer solution Comment 1 of 1
72 out of the 362880 permutations of the nine digits satisfy the criteria for a probability of 1/5040 ~= 0.000198412698412698.

Filtering out rotations and reflections by requiring the top-left number to be the smallest corner and the top-right number to be smaller than the bottom-left number, the following are the nine basic ways the criteria can be met:

126
837
945

153
297
648

153
864
972

216
738
954

243
198
657

315
864
729

324
198
576

324
765
819

657
432
819

DefDbl A-Z
Dim crlf\$, dig(9)

Form1.Visible = True

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

d\$ = "123456789": h\$ = d\$
Do
DoEvents
For i = 1 To Len(d)
dig(i) = Val(Mid(d, i, 1))
Next
good = 1
For maj = 0 To 6 Step 3
tot = 0
For mnr = 1 To 3
tot = tot + dig(maj + mnr)
Next
If tot Mod 9 <> 0 Then good = 0: Exit For
Next
If good Then
For maj = 1 To 3
tot = 0
For mnr = 0 To 6 Step 3
tot = tot + dig(maj + mnr)
Next
If tot Mod 9 <> 0 Then good = 0: Exit For
Next
If good Then
tot1 = dig(1) + dig(5) + dig(9)
tot2 = dig(3) + dig(5) + dig(7)
If tot1 Mod 9 = 0 And tot2 Mod 9 = 0 Then
If dig(1) < dig(3) And dig(3) < dig(7) And dig(1) < dig(9) Then
Text1.Text = Text1.Text & Mid(d, 1, 3) & crlf
Text1.Text = Text1.Text & Mid(d, 4, 3) & crlf
Text1.Text = Text1.Text & Mid(d, 7, 3) & crlf
Text1.Text = Text1.Text & crlf
End If
ct = ct + 1
End If
End If
End If

overct = overct + 1

permute d
Loop Until d = h

Text1.Text = Text1.Text & ct & Str(overct) & Str(ct / overct) & crlf & " done"

End Sub

 Posted by Charlie on 2015-10-26 10:55:37

 Search: Search body:
Forums (0)