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1248 must be included (Posted on 2016-09-06) Difficulty: 2 of 5
There are three infinite arithmetic progressions.
Given that each of the numbers 1,2,3,4,5,6,7,8 occurs at least in one of these progressions, show that the number 1248 must occur at least in one of them.

No Solution Yet Submitted by Ady TZIDON    
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Solution Brute force (computer solution) | Comment 1 of 3
Checking every partition into three groups,  (6561 partitions counted as checked, which is 3^8):

DefDbl A-Z
Dim crlf$, prg$(3)


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For one = 1 To 3
   prg(one) = prg(one) + "1"
 For two = 1 To 3
   prg(two) = prg(two) + "2"
 For three = 1 To 3
   prg(three) = prg(three) + "3"
 For four = 1 To 3
   prg(four) = prg(four) + "4"
 For five = 1 To 3
   prg(five) = prg(five) + "5"
 For six = 1 To 3
   prg(six) = prg(six) + "6"
 For seven = 1 To 3
   prg(seven) = prg(seven) + "7"
 For eight = 1 To 3
   prg(eight) = prg(eight) + "8"
   
   partitions = partitions + 1
   found = 0
   For p = 1 To 3
     If Len(prg(p)) > 1 Then
       If Len(prg(p)) = 2 Then
         If p = 2 Then
           x = x
         End If
         diff = Val(Right(prg(p), 1)) - Val(Left(prg(p), 1))
         If diff <= 0 Then
           Text1.Text = Text1.Text & "PROGRAM ERROR"
           Exit Sub
         End If
         diffmajor = 1248 - Val(Left(prg(p), 1))
         If diffmajor Mod diff = 0 Then
           found = 1
           Exit For
         End If
       Else
        diff = Val(Mid(prg(p), 2, 1)) - Val(Left(prg(p), 1))
        For i = 3 To Len(prg(p))
          diff2 = Val(Mid(prg(p), i, 1)) - Val(Mid(prg(p), i - 1, 1))
         If diff <= 0 Or diff2 <= 0 Then
           Text1.Text = Text1.Text & "PROGRAM ERROR"
           Exit Sub
         End If
          diff = gcd(diff, diff2)
        Next
        diffmajor = 1248 - Val(Left(prg(p), 1))
       End If
       If diffmajor Mod diff = 0 Then
           found = 1
           Exit For
       End If
     End If
   Next
   If found = 0 Then
    For j = 1 To 3
     Text1.Text = Text1.Text & prg(j) & " "
    Next
    Text1.Text = Text1.Text & crlf
   End If
   
   ix = InStr(prg(eight), "8")
   prg(eight) = Left(prg(eight), ix - 1) + Mid(prg(eight), ix + 1)
 Next
   ix = InStr(prg(seven), "7")
   prg(seven) = Left(prg(seven), ix - 1) + Mid(prg(seven), ix + 1)
 Next
   ix = InStr(prg(six), "6")
   prg(six) = Left(prg(six), ix - 1) + Mid(prg(six), ix + 1)
 Next
   ix = InStr(prg(five), "5")
   prg(five) = Left(prg(five), ix - 1) + Mid(prg(five), ix + 1)
 Next
   ix = InStr(prg(four), "4")
   prg(four) = Left(prg(four), ix - 1) + Mid(prg(four), ix + 1)
 Next
   ix = InStr(prg(three), "3")
   prg(three) = Left(prg(three), ix - 1) + Mid(prg(three), ix + 1)
 Next
   ix = InStr(prg(two), "2")
   prg(two) = Left(prg(two), ix - 1) + Mid(prg(two), ix + 1)
 Next
  ix = InStr(prg(one), "1")
   prg(one) = Left(prg(one), ix - 1) + Mid(prg(one), ix + 1)
 Next
 
 Text1.Text = Text1.Text & crlf & partitions & " done"
  
End Sub

Function gcd(a, b)
  x = a: y = b
  Do
   q = Int(x / y)
   z = x - q * y
   x = y: y = z
  Loop Until z = 0
  gcd = x
End Function


No sets were found where no partition formed a progression without 1248.

  Posted by Charlie on 2016-09-06 10:19:42
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