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Double perfection (Posted on 2018-12-28) Difficulty: 3 of 5
Find the smallest integer for which both the number of its divisors and the sum of its prime factors are perfect numbers.

Rem: Duplicate primes are summed up: e.g. 12 = 2*2*3, - so the sum of 12's prime factors is 2+2+3=7.

Bonus question: Any others?

See The Solution Submitted by Ady TZIDON    
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Solution computer solution Comment 1 of 1
338 has 6 divisors (1, 2, 13, 26, 169 and 338) and the sum of its prime factors is 28 (2+13+13).

Two other such numbers are 2496 and 3520.

 338  6 28
2496 28 28
3520 28 28

DefDbl A-Z
Dim crlf$
Dim fct(20, 1), perfect As Variant
Private Sub Form_Load()
t = Timer
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
  
 perfect = Array(0, 6, 28, 496, 8128, 33550336, 8589869056#, 137438691328#)

 For n = 2 To 50000
   f = factor(n)
   totpf = 0
   For i = 1 To f
     totpf = totpf + fct(i, 0) * fct(i, 1)
   Next i
   good = 0
   For i = 1 To 7
     If totpf = perfect(i) Then good = 1: Exit For
   Next i
   If good Then
    numdiv = 1
    For i = 1 To f
      numdiv = numdiv * (fct(i, 1) + 1)
    Next i
    good = 0
    For i = 1 To 7
     If numdiv = perfect(i) Then
       good = 1: Exit For
     End If
    Next i
    If good Then
      Text1.Text = Text1.Text & n & Str(numdiv) & Str(totpf) & crlf
    End If
   End If
   DoEvents
 Next n

  Text1.Text = Text1.Text & " end"

End Sub

Function factor(num)
 diffCt = 0: good = 1
 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
   If INKEY$ = Chr$(27) Then s$ = Chr$(27): Exit Function
 Loop
 If n > 1 Then diffCt = diffCt + 1: fct(diffCt, 0) = n: fct(diffCt, 1) = 1
 factor = diffCt
 Exit Function

DivideIt:
 cnt = 0
 Do
  q = Int(n / dv)
  If q * dv = n And n > 0 Then
    n = q: cnt = cnt + 1: If n > 0 Then limit = Sqr(n) Else limit = 0
    If limit <> Int(limit) Then limit = Int(limit + 1)
   Else
    Exit Do
  End If
 Loop
 If cnt > 0 Then
   diffCt = diffCt + 1
   fct(diffCt, 0) = dv
   fct(diffCt, 1) = cnt
 End If
 Return
End Function


  Posted by Charlie on 2018-12-28 15:26:08
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