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4 partitions, same product (Posted on 2016-03-31) Difficulty: 3 of 5
What is the smallest integer that has exactly 4 different partitions into 3 parts with the same product?

a. List these partitions.
b. (optional) Find the second smallest number displaying the above feature.
c. Bonus: Why the specification "exactly"?

No Solution Yet Submitted by Ady TZIDON    
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Solution computer solution and discussion Comment 1 of 1
The following program backs into the solution by finding all the ways of factoring any number, n, into three factors to find one where the totals of the three factors are identical in four cases.

DefDbl A-Z
Dim crlf$, tots(), cts()

Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For n = 4 To 2460375
   ReDim tots(4000), cts(4000)
   cr = Int(n ^ (1 / 3) + 0.5)
   For f1 = 1 To cr
     DoEvents
     n2 = n / f1
     If n2 = Int(n2) Then
       For f2 = f1 To Int(Sqr(n2) + 0.5)
         n3 = n2 / f2
         If n3 = Int(n3) Then
           f3 = n3
           tot = f1 + f2 + f3
           found = 0
           For i = 1 To 4000
             If tots(i) = 0 Then Exit For
             If tots(i) = tot Then
               found = 1
               cts(i) = cts(i) + 1
               Exit For
             End If
           Next i
           If found = 0 Then
               tots(i) = tot
               cts(i) = 1
           End If
         End If
       Next f2
     End If
   Next f1
   For i = 1 To 4000
     If cts(i) >= 4 And tots(i) <= 135 Then
       Text1.Text = Text1.Text & n & Str(tots(i)) & Str(cts(i)) & crlf
     End If
   Next
 Next n
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

The results are

          how many
prod. sum tri-partitions
32760 130 4
37800 118 4
45360 135 4
50400 133 4
60480 130 4
 done
 
where the factoring was done on numbers up to 135^3.  Numbers higher than this would have factors totalling more than 135.

None were found where there were more than four ways of getting a total and a product to agree, having the total = product <= 135.

The partitions of 118 that multiply to 37800 are:
14 50 54
15 40 63
18 30 70
21 25 72

Now, 130 can be partitioned in exactly four ways so that the partitions multiply out to 32760.  But it can also be partitioned in exactly four ways to so these partitions multiply out to 60480. 

130 for 32760
-------------
9 56 65
10 42 78
14 26 90
15 24 91

130 for 60480
-------------
20 54 56
21 45 64
24 36 70
28 30 72

I don't think the presence of these two sets of partitionings violate the "exactly 4" rule, as each set independently give exactly 4 for the same product.

But in case that would be disqualified we still have:

For 133 producing 50400:
15 48 70
16 42 75
18 35 80
24 25 84

and 135 producing 45360:
12 60 63
14 40 81
15 36 84
21 24 90

The auxilliary program for these (to list the actual partitions) is:

DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 tot = 135: prod = 45360
 Text1.Text = tot & Str(prod) & crlf
 For a = 1 To tot / 3
 For b = a To (tot - a) / 2
   c = tot - a - b
   If a * b * c = prod Then Text1.Text = Text1.Text & a & Str(b) & Str(c) & crlf
 Next
 Next
  
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

with variations in the line

 tot = 135: prod = 45360


  Posted by Charlie on 2016-03-31 10:32:29
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