The math rule is that if a sum of the digits of a number can be divided by 3 without remainder, than the number is self can be divided by 3 without remainder. I don't know if this has been proven, but I know for sure that it works. So if you calculate the sum of the digits of a number and it's 3, 6 or 9, you can be sure it can be divided by 3 without remainder. The function I wrote below just sums the digits recursively until there is a single digit. You can do the final if check on your own :D
Function DigitSum(txt As String) As String Dim rez As Long
Dim srez As String
rez = 0
'tt = ""
For i = 1 To Len(txt)
k = Mid(txt, i, 1)
'tt = tt & k
j = CInt(k)
rez = rez + j
Next i
srez = CStr(rez)
While (Len(srez) > 1)
srez = DigitSum(srez)
Wend
DigitSum = srez
End Function
