The solution to your specific problem has been identified already.

This is an interesting problem. The number of digits in the integer part of
any number is given by:

int(1+log(YourNumber)/log(10))

where YourNumber can be an integer or float. Say YourNumber is 1236.45, the
last digit is:

Debug.Print mid(1236.45,int(1+log(1236.45)/log(10)),1)

This can be extended; say, you wanted the last but one digit, the position is:

int(1+log(YourNumber)/log(10)) -1

etc..