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Consecutive Conclusion (Posted on 2014-09-30) Difficulty: 3 of 5
Determine the maximum possible value of N for which 311 is expressible as the sum of N consecutive positive integers.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

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Solution two computer algorithm solutions | Comment 6 of 8 |
The first version uses a formula for first element based on n, which must be a power of 3 or twice a power of 3:

DefDbl A-Z
Dim crlf$, used(25)


Private Sub Form_Load()
 ChDir "C:Program Files (x86)DevStudioVBprojects looble"
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 DoEvents
 num = 2 * Int(3 ^ 11 + 0.5)
 n1 = 3: n2 = 6
 Do
  prevn1 = n1: prevn2 = n2
  n1 = n1 * 3: n2 = n2 * 3
  a1 = (num / n1 - n1 + 1) / 2
  a2 = (num / n2 - n2 + 1) / 2
  If n1 > maxn And a1 > 0 And a1 = Int(a1) Then maxn = n1: maxa = a1
  If n2 > maxn And a2 > 0 And a2 = Int(a2) Then maxn = n2: maxa = a2
 Loop Until a1 < 1 And a2 < 1
 Text1.Text = Text1.Text & maxn & Str(maxa) & crlf
 
 tot = 0
 For i = maxa To maxa + maxn - 1
  tot = tot + i
 Next
 Text1.Text = Text1.Text & tot & Str(num / 2) & Str(maxn * (2 * maxa + maxn - 1) / 2) & crlf
 
 Text1.Text = Text1.Text & crlf & " done"
End Sub

finds

486 122
177147 177147 177147

which indicates the maximum N is 486, when the arithmetic seriest begins at 122.

The three copies of 177147 indicated the calculated sum, a repetition of the calculation 3^11 and the actual sum given by adding the 486 elements, as a check.


while the second method starts with the lowest arithmetic series and increases or decreases it in order to get to 3^11:

DefDbl A-Z
Dim crlf$, used(25)


Private Sub Form_Load()
 ChDir "C:Program Files (x86)DevStudioVBprojects looble"
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 DoEvents
 num = Int(3 ^ 11 + 0.5)
 First = 1: Last = 1: tot = 1
 Do
  If tot < num Then
    Last = Last + 1
    tot = tot + Last
  ElseIf tot > num Then
    tot = tot - First
    First = First + 1
  Else
    Exit Do
  End If
 Loop
 Text1.Text = Text1.Text & First & Str(Last) & "           " & Last - First + 1 & crlf

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


works to find the lowest starting point, which will result in the highest N. It finds:

122 607           486

indicating the series starts at 122 and ends at 607 for N = 486.

A change to the first version will show all the arithmetic series with increment 1 that find 3^11:

 n   first 
    element
  9 19679
 18  9833
 27  6548
 54  3254
 81  2147
162  1013
243   608
486   122

while the change to the second version finds more:

 first   last           n
    122    607         486
    608    850         243
   1013   1174         162
   2147   2227          81
   3254   3307          54
   6548   6574          27
   9833   9850          18
  19679  19687           9
  29522  29527           6
  59048  59050           3
  88573  88574           2
 177147 177147           1

The reason the first didn't find the extra, is that it started searching for odd n at n=9, and for even n at n=18, on the basis that the maximum n would be higher, and it was.

Edited on October 1, 2014, 7:42 am
  Posted by Charlie on 2014-09-30 18:40:42

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