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240 divisors (Posted on 2016-08-15) Difficulty: 2 of 5
Find the smallest positive integer that has precisely 240 positive divisors.

No Solution Yet Submitted by Ady TZIDON    
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Solution Analytic guesses plus computer verification (spoiler) Comment 1 of 1
240 = 2^4 * 3 * 5

That could be 2 * 2 * 2 * 2 * 3 * 5 which could imply the number sought was

2^4 * 3^2 * 5 * 7 * 11 * 13 = 720,720

(Each factor of 2 in 240 implies a prime factor to the 1st power; a factor of 3 in 240 implies a prime factor squared; the factor of 5 in 240 implies a prime factor to the 4th power.)

There could be alternate factorings of 240, such as 2 * 2 * 3 * 4 * 5, which would correspond to the number's being

2^4 * 3^3 * 5^2 * 7 * 11 = 831,600

Clearly the  720,720 is the lower of the two and in fact this program verifies it's the lowest of such reworkings:

DefDbl A-Z
Dim crlf$, fct(20, 1)


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For n = 1 To 1000000
   f = factor(n)
   divisors = 1
   For i = 1 To f
     divisors = divisors * (fct(i, 1) + 1)
   Next
   If divisors = 240 Then
     Text1.Text = Text1.Text & n & crlf
       For i = 1 To f
         Text1.Text = Text1.Text & "     " & fct(i, 0) & Str(fct(i, 1)) & crlf
       Next
   End If
   DoEvents
 Next
 
 Text1.Text = Text1.Text & crlf & " done"
  
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


The output being the following, where the valid number with 240 divisors is on the first line of each group followed by a list of its prime factors, with the power to which that factor is raised.

Second place is indeed the 

831,600 = 2^4 * 3^3 * 5^2 * 7 * 11

we found above, resulting from a factoring of 240 = 5 * 4 * 3 * 2 * 2



720720
     2 4
     3 2
     5 1
     7 1
     11 1
     13 1
831600
     2 4
     3 3
     5 2
     7 1
     11 1
942480
     2 4
     3 2
     5 1
     7 1
     11 1
     17 1
982800
     2 4
     3 3
     5 2
     7 1
     13 1
997920
     2 5
     3 4
     5 1
     7 1
     11 1


  Posted by Charlie on 2016-08-15 10:27:39
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