
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 2digit
remainders (10, 11, 12, 13, 14, 15) replaces these values with
alphabetic representations (10A, 11B, 12C, 13D, 14E, 15F).
The answer is:
110_{10 }
= 6E_{16}
(a hexadecimal
conversion) 