The process remains the same. BUT there is one
problem in base 16 that did not appear in the examples above. One
of the remainders in this division contains 2 digits (14). You
CANNOT allow 2 digits to reside in one of the place holdings in a
number. For this reason, base 16, which can have six 2-digit
remainders (10, 11, 12, 13, 14, 15) replaces these values with
alphabetic representations (10-A, 11-B, 12-C, 13-D, 14-E, 15-F).
The answer is: