Number systems: binary, denary and hexadecimal conversion, binary addition, two's complement for signed numbers, and binary-coded decimal.

Binary (up to 3 marks): subtract place values $128, 64, 32, 16, 8, 4, 2, 1$. $173 - 128 = 45$ (bit set), $45 - 32 = 13$ (bit set, $64$ not), $13 - 8 = 5$, $5 - 4 = 1$, $1 - 1 = 0$. So the set bits are $128, 32, 8, 4, 1$: $173 = 1010\,1101_2$.

Hexadecimal (up to 2 marks): group the binary into nibbles, $1010\,1101$, and convert each: $1010 = \text{A}$, $1101 = \text{D}$. So $173 = \text{AD}_{16}$ (check: $10 \times 16 + 13 = 173$).

Markers reward the correct 8-bit binary $10101101$ and the hexadecimal $\text{AD}$, with the place-value or nibble method shown.

Two's complement of $-20$ (up to 3 marks): write $+20 = 0001\,0100$. Invert all bits to get $1110\,1011$, then add $1$: $1110\,1100$. So $-20 = 1110\,1100$ in 8-bit two's complement.

Subtraction by addition (up to 3 marks): $35 = 0010\,0011$. Add $-20$: $0010\,0011 + 1110\,1100 = 1\,0000\,1111$. Discard the carry out of the 8th bit, leaving $0000\,1111 = 15$. So $35 - 20 = 15$.

Markers reward $-20 = 11101100$ (invert and add one), the binary addition, discarding the final carry, and the answer $15$.