At most six primes can be formed using each of the ten digits exactly once, as each prime must end with an odd digit or be 2. In addition to one of the primes having to be 2, another must be 5 as any number ending in 5 is not prime unless the number is 5 itself.

The program reports the 2 and the 5 as well as the other four primes that can be formed from such partitioning.

DefDbl A-Z

Dim pr(4), crlf$

Private Sub Form_Load()

 Text1.Text = ""

 crlf$ = Chr(13) + Chr(10)

 Form1.Visible = True

 a$ = "13467890": h$ = a$

 Do

   If Left(a$, 1) <> "0" Then

    If InStr("1379", Right(a$, 1)) > 0 Then

     a2$ = a$

     pct = 0

     good = 1

     Do While a2$ > ""

        For ix = 1 To Len(a2$)

          If InStr("1379", Mid(a2$, ix, 1)) > 0 Then

            pct = pct + 1

            pr(pct) = Val(Left(a2$, ix))

            If Left(a2$, 1) = "0" Then good = 0: Exit For

            a2$ = Mid(a2$, ix + 1)

            Exit For

          End If

        Next

        If good = 0 Then Exit Do

     Loop

     For i = 1 To 4

        If prmdiv(pr(i)) <> pr(i) Or pr(i) = 1 Then good = 0: Exit For

        If pr(i) < pr(i - 1) Then good = 0: Exit For

     Next

     

     If good Then

       flag = 0

       For i = 1 To 4

         If pr(i) > Max Then flag = 1: Max = pr(i)

       Next

       Text1.Text = Text1.Text & "2 5 "

       For i = 1 To 4

         Text1.Text = Text1.Text & Str(pr(i))

       Next

       If flag Then Text1.Text = Text1.Text & " ***"

       Text1.Text = Text1.Text + crlf$

       DoEvents

     End If

    End If

   End If

   permute a$

 Loop Until a$ = h$

End Sub

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

finds 13 ways of partitioning the 10 digits into six primes:

2 5  3 41 67 809 ***

2 5  3 41 89 607

2 5  3 47 61 809

2 5  3 47 89 601

2 5  3 67 89 401

2 5  3 7 41 6089 ***

2 5  3 7 41 8069 ***

2 5  3 7 41 8609 ***

2 5  3 7 461 809

2 5  3 7 641 809

2 5  7 43 61 809

2 5  7 43 89 601

2 5  7 61 83 409

So in one interpretation of the puzzle, any one of these lines can be used as the partitioning, and the rightmost number on the line is the largest prime created from that partitioning.

But to get a unique answer you can take a more restrictive interpretation of the last sentence: that the partitioning is to be done in such a way as to include the highest possible prime. To facilitate that, the *** in the above listing marks a partitioning that includes a prime higher than any prime in the list up to that point. The last such marked partitioning is

2 5  3 7 41 8609

and therefore 8609 is the highest prime that can be created thus.