For being hexadecimal 314086807350 looks like a "special" code though, since there is no 2 or 9 neither are there are any letters, A, B, C, D, E or F present.
But if we simply asume it is valid hexadecimal number, with each figure being an hexdigit or hexit, values represented from 0 to F.
Then there is a technique to add a check sum, I do not know where it is supposed to go. At the left (*a* it is 31) or at the right (*b* is it 50)
Acording to Wikipedia the calculation of the hexadecimal check sum requires several steps:
1) Add the hexadecimal digits
*a* 31+40+86+80+73 = 1EA
*b* 40+86+80+73+50 = 209
here we did hexadecimal arithmetic and the couple of 8 and absence of letters helped!
2) dropping the carry nibble, always a simple operation
3) get its two's complement EDIT this line was missing in the original post sorry *a* becomes EA <> 50
*b* becomes 09 <> 31
so the senidenary check sum is NOT correct in any case.
So even if we do not know in which case, *a* or *b* we are, we have to ask the Central Processing Unit to send the code again since we have received an invalid information (here is my favorite part: because there might be some malfunction in the Multiplexor, or some digital compenent, or some magnetic field distorted the code during the transmition, or some virus got its hand on it or a bug in the software created an invalid code or... or...) whatever the cause, no Arithmetic and Logic Unit would ever accept to process "corrupted" codes.