Convert the following binary numbers into their hexadecimal equivalents:
(a) $11001111_{2}$
(b) $110011110_{2}$
(a) Grouping bits in fours from the right gives: $\underbrace{1100}_{\mathrm{C}} \underbrace{11 \|}_{\mathrm{F}}$ and assigning hexadecimal symbols to each group gives as above, from Table $10.2$.
Thus, $11001111_{2}=\mathbf{C F}_{16}$
(b) Grouping bits in fours from the right gives: $\underbrace{0001}_{1} \underbrace{1001}_{9} \underbrace{1110}_{\mathbf{E}}$ and assigning hexadecimal symbols to each group gives as above, from Table $10.2$.
Thus, $110011110_{2}=19 \mathbb{E}_{16}$
(d) Converting from hexadecimal to binary
The above procedure is reversed; thus, for example,
$$
6 \mathrm{CF} 3_{16}=0110110011110011
$$
from Table $10.2$
i.e. $6 \mathrm{CF} 3_{16}=110110011110011_{2}$