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3 sets of primes (Posted on 2015-03-08) Difficulty: 4 of 5
What is the smallest integer that can be expressed as a sum of consecutive primes in 3 different ways?

See The Solution Submitted by Ady TZIDON    
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Solution computer solution Comment 1 of 1
The program (at the bottom) was designed to use only up to the 50th prime. That means that in order to insure completeness for all reported numbers, those numbers should not exceed 456, as the sum of the 49th and 50th primes.  Also, as a further consequence, only a running count for up to 18 primes would be necessary, as 19 or more consecutive primes would add up to more than 456.

The intended answer is probably 240, as:

240 = 17+19+23+29+31+37+41+43
= 53+59+61+67
= 113+127

This is reported as the entry below:

240
7 14:   17 19 23 29 31 37 41 43
16 19:   53 59 61 67
30 31:   113 127

where the numbers before the colon refer to the ordinality of the primes that follow; for example 17 is the 7th prime and 43 is the 14th.

287 would be the next such number.

In the entries below, those in which the target number is also prime are marked with **. That's to indicate that the number itself could be considered a degenerate "sum" of consecutive primes. If that were done, 41 would be the lowest, as there are two other sums of consecuteve primes that add up to 41.

Interestingly 311 has four sums of actual consecutive primes that add up to it, plus it is itself prime, giving a 5th, degenerate, sum.

41 **
1 6:   2 3 5 7 11 13
5 7:   11 13 17

83 **
5 9:   11 13 17 19 23
9 11:   23 29 31

197 **
1 12:   2 3 5 7 11 13 17 19 23 29 31 37
7 13:   17 19 23 29 31 37 41

199 **
11 15:   31 37 41 43 47
18 20:   61 67 71

223 **
8 14:   19 23 29 31 37 41 43
20 22:   71 73 79

240
7 14:   17 19 23 29 31 37 41 43
16 19:   53 59 61 67
30 31:   113 127

251 **
9 15:   23 29 31 37 41 43 47
22 24:   79 83 89

281 **
1 14:   2 3 5 7 11 13 17 19 23 29 31 37 41 43
10 16:   29 31 37 41 43 47 53

287
7 15:   17 19 23 29 31 37 41 43 47
15 19:   47 53 59 61 67
24 26:   89 97 101

311 **
5 15:   11 13 17 19 23 29 31 37 41 43 47
11 17:   31 37 41 43 47 53 59
16 20:   53 59 61 67 71
26 28:   101 103 107

340
7 16:   17 19 23 29 31 37 41 43 47 53
10 17:   29 31 37 41 43 47 53 59
39 40:   167 173

371
4 16:   7 11 13 17 19 23 29 31 37 41 43 47 53
13 19:   41 43 47 53 59 61 67
30 32:   113 127 131

401 **
10 18:   29 31 37 41 43 47 53 59 61
14 20:   43 47 53 59 61 67 71

439 **
11 19:   31 37 41 43 47 53 59 61 67
34 36:   139 149 151

DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True

 hiPrime = 50
 maxGoal = prm(hiPrime - 1) + prm(hiPrime)
 maxCt = 0
 While tot <= maxGoal
   maxCt = maxCt + 1
   tot = tot + prm(maxCt)
 Wend
 Text1.Text = Text1.Text & maxCt & Str(maxGoal) & crlf
 ReDim totPrm(maxCt), ways(maxGoal, 10, 2)
 For p = 1 To hiPrime
   For i = 2 To maxCt
     totPrm(i) = totPrm(i) + prm(p)
     prevP = p - i
     If prevP > 0 Then
      totPrm(i) = totPrm(i) - prm(prevP)
     End If
     If p >= i Then
       goal = totPrm(i)
       If goal <= maxGoal Then
         ways(goal, 0, 0) = ways(goal, 0, 0) + 1
         s = ways(goal, 0, 0)
         ways(goal, s, 1) = prevP + 1
         ways(goal, s, 2) = p
       End If
     End If
   Next
 Next p
 
 For i = 1 To maxGoal
   If ways(i, 0, 0) > 2 Or prmdiv(i) = i And ways(i, 0, 0) > 1 Then
    Text1.Text = Text1.Text & i
    If prmdiv(i) = i Then Text1.Text = Text1.Text & " **"
    Text1.Text = Text1.Text & crlf
    For j = 1 To ways(i, 0, 0)
     Text1.Text = Text1.Text & ways(i, j, 1) & Str(ways(i, j, 2)) & ":  "
     For k = ways(i, j, 1) To ways(i, j, 2)
      Text1.Text = Text1.Text & Str(prm(k))
     Next
     Text1.Text = Text1.Text & crlf
    Next
    Text1.Text = Text1.Text & crlf
   End If
 Next
 

End Sub

Function prm(i)
  Dim p As Long
  Open "17-bit primes.bin" For Random As #111 Len = 4
  Get #111, i, p
  prm = p
  Close 111
End Function

Function prmdiv(num)
 Dim n, dv, q
 If num = 1 Then prmdiv = 1: Exit Function
 n = Abs(num): If n > 0 Then limit = Sqr(n) Else limit = 0
 If limit <> Int(limit) Then limit = Int(limit + 1)
 dv = 2: GoSub DivideIt
 dv = 3: GoSub DivideIt
 dv = 5: GoSub DivideIt
 dv = 7
 Do Until dv > limit
   GoSub DivideIt: dv = dv + 4 '11
   GoSub DivideIt: dv = dv + 2 '13
   GoSub DivideIt: dv = dv + 4 '17
   GoSub DivideIt: dv = dv + 2 '19
   GoSub DivideIt: dv = dv + 4 '23
   GoSub DivideIt: dv = dv + 6 '29
   GoSub DivideIt: dv = dv + 2 '31
   GoSub DivideIt: dv = dv + 6 '37
 Loop
 If n > 1 Then prmdiv = n
 Exit Function

DivideIt:
 Do
  q = Int(n / dv)
  If q * dv = n And n > 0 Then
    prmdiv = dv: Exit Function
   Else
    Exit Do
  End If
 Loop

 Return
End Function

  Posted by Charlie on 2015-03-08 12:53:52
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