Hi Poco90,
The problem has nothing to do with numbers vs. characters, but rather in the fact that you appear to be using 16-digit part numbers rather than the 15 you mentioned in your problem statement. Either that, or the zero at the end of each of your strings is the "extra random digit" you mentioned. If the zero at the end is truly part of the part number you want to match and therefore some of the part numbers are 16 digits long, you must use 16 in the corresponding MID function, like this:
=IF(OR(MID(A1,1,16)="MNRIE14E0ABEJ220",MID(A1,1,16)="MNRIE14E0AB J220",MID(A1,1,15)="MCRIE03EZHEJ3R0",MID(A1,1,15)="MCRIE03EZPEJ3R0",MID(A1,1,15)="MCRIE03EZH J3R0"),"Yes","No")
where the last three are compared to 15 characters because they are 15 characters long.
If, on the other hand, the zero of the 16-character part numbers IS NOT part of the part number you want to compare (is an extra random digit), then do the compare against the part numbers without the zeros:
=IF(OR(MID(A1,1,15)="MNRIE14E0ABEJ22",MID(A1,1,15)="MNRIE14E0AB J22",MID(A1,1,15)="MCRIE03EZHEJ3R0",MID(A1,1,15)="MCRIE03EZPEJ3R0",MID(A1,1,15)="MCRIE03EZH J3R0"),"Yes","No")
Note that the space does count as a character, as you presumed.
You also mentioned that MNRIE14E0ABEJ220 scans as MNRIE14E0ABJ220 J, which not only has an extra two characters at the end, but also is missing the E. If the " J" is what you refer to as an extra random character, the MID function in the formula should get rid of it, but the missing E is a problem. To match this you should have
MID(A1,1,15)="MNRIE14E0ABJ220"
in the formula, i.e., comparing it to the part number without the E.
Does this help?
Damon