Split cells into 2 columns when any number - vba

Question

I can see numerous posts around this topic but none that specifically solves the problem I have.
I have a string that has text and numbers. I need to split the string into 2 columns when it first sees a number.
Example:
Ballyvic Boru5/6
First Drift2/1
Sizing Cusimanoin15/2
Becomes:

Answer 1

You can use a simple formula to find the first number, along with LEFT and MID to split the string.
Part 1:
=LEFT(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890"))-1)
Part 2:
=MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),99)

Answer 2

Here's a regex method:
You must set a reference to Microsoft VBScript Regular Expressions x.x, where x.x is the highest version you have (mine is 5.5)
Option Explicit
Sub splitCells()
Dim RegEx As New RegExp, i As Long, tempStr As String
With RegEx
.Global = True
.IgnoreCase = True
.Pattern = "(([a-z]*\s?)*\D)(\d.*)"
End With
With ThisWorkbook.Worksheets(1)
For i = 1 To .Cells(.Rows.Count, 1).End(xlUp).Row
If RegEx.Test(.Cells(i, 1)) Then
tempStr = .Cells(i, 1)
.Cells(i, 1) = RegEx.Replace(tempStr, "$1")
.Cells(i, 2) = RegEx.Replace(tempStr, "$3")
End If
Next i
End With
End Sub
Breaking down the Regular Expression:
(([a-z]*\s?)*\D)(\d.*)
[a-z]* matches any character in the alphabet, with the * multiplier for unlimited occurances
\s? Matches any whitespace character, with the ? multiplier to match 0-1 occurances (meaning there may or may not be a white space
Both of the above is enclosed in a grouping (), followed by another * to match 0-unlimited occurances
\D This excludes all digits
The above is enclosed in a group with the first (([..])*\D)
We have our final group: (\d.*), which matches the first digit and everything else afterwards.

Answer 3

Here's a pair functions you can use on the worksheet (as opposed to having to run a VBA procedure to 'fix' the cells one time):
Public Function splitNum1(str As String) As String
Dim p
For p = 1 To Len(str)
If Mid(str, p, 1) Like "#" Then Exit For
Next
splitNum1 = Left(str, p - 1)
End Function
Public Function splitNum2(str As String) As String
splitNum2 = Right(str, Len(str) - Len(splitNum1(str)))
End Function
splitNum1 returns the string on the "left" side of the number.
splitNum2 returns the string beginning with the first nummber.