Number Systems — worked example

Convert $173_{10}$ into binary, octal, hex, BCD, and Gray code. Each base reuses the same value, and the conversions all go through the binary representation.

1. Binary by repeated division. Divide by two; record the remainder; feed the quotient back. Read the remainders bottom-to-top.
   $$\begin{aligned}173 &\div 2 = 86 \;\text{r}\; 1 \\ 86 &\div 2 = 43 \;\text{r}\; 0 \\ 43 &\div 2 = 21 \;\text{r}\; 1 \\ 21 &\div 2 = 10 \;\text{r}\; 1 \\ 10 &\div 2 = 5 \;\text{r}\; 0 \\ 5 &\div 2 = 2 \;\text{r}\; 1 \\ 2 &\div 2 = 1 \;\text{r}\; 0 \\ 1 &\div 2 = 0 \;\text{r}\; 1\end{aligned}$$
   Remainders bottom-up: $173_{10} = 10101101_{2}$.
2. Hex by regrouping. Split the 8-bit binary into two nibbles, then translate each nibble (0..15) into one hex digit (0..F).
   $$\underbrace{1010}_{A} \;\; \underbrace{1101}_{D} \;\Rightarrow\; AD_{16}$$
   Cross-check: $A \cdot 16 + D = 10 \cdot 16 + 13 = 173$.
3. Octal by regrouping. Pad the binary with a leading zero to a multiple of three bits, then split into 3-bit groups.
   $$\underbrace{0\,10}_{2} \;\; \underbrace{101}_{5} \;\; \underbrace{101}_{5} \;\Rightarrow\; 255_{8}$$
   Cross-check: $2 \cdot 64 + 5 \cdot 8 + 5 = 128 + 40 + 5 = 173$.
4. BCD digit-by-digit. Encode each decimal digit of $173$ independently in 4 bits.
   $$1 \to 0001 \quad 7 \to 0111 \quad 3 \to 0011$$
   $$\therefore 173_{10} = 0001\,0111\,0011_{\text{BCD}}$$
   Note that BCD is wider than straight binary: 12 bits versus 8 bits. The trade-off is human-readable digits at the cost of bit density.
5. Gray by XOR. Apply $g = n \oplus (n \gg 1)$ bit-wise.
   $$\begin{aligned}n &= 1010\,1101 \\ n \gg 1 &= 0101\,0110 \\ g &= 1111\,1011\end{aligned}$$
   So $173_{10} = 11111011_{\text{Gray}}$. The XOR collapses to one Gray bit-flip per increment in $n$, which is the entire point of the encoding.