Exercises — Hexadecimal and octal representation

Hex borrows letters once it runs out of $0$–$9$: $A=10,B=11,C=12,D=13,E=14,F=15$. So $\text{C}=12$ and $\text{E}=14$.

"$10$" means $1\cdot b^1+0\cdot b^0=b$. So it equals the base itself. $10_2=2$, $10_8=8$, $10_{16}=16$.

Octal digits run $0$–$7$ only. A digit $\ge 8$ is illegal. Legal: $47_8$ (4,7 both ≤7) and $17_8$. Illegal: $58_8$ (has an 8), $80_8$ (has an 8).

Two hex digits → place values $16^1$ and $16^0$. And $\text{A}=10$. $3A_{16}=3\cdot 16^1+10\cdot 16^0=48+10=58_{10}.$

Three octal digits → place values $8^2,8^1,8^0$. $264_8=2\cdot 64+6\cdot 8+4\cdot 1=128+48+4=180_{10}.$

Divide by 16, keep remainders. Why: the remainder of $N÷b$ is exactly the last digit $d_0$, because every higher term $d_i{b}^i$ is a multiple of $b$. $180÷16=11\text{ r }411÷16=0\text{ r }11(=\text{B})$ Read remainders bottom→top (first remainder is the $16^0$ place, so it goes rightmost): $\text{B},4\Rightarrow \mathbf{B}{4}_{16}$. Check: ${\text{B4}}_{16}=11\cdot 16+4=176+4=180$. ✓

Hex — group into 4s from the right: $\underset{\text{B}}{\underbrace{1011}}\underset{5}{\underbrace{0101}}$. $1011_2=8+2+1=11=\text{B}$, and $0101_2=5$. So $\mathbf{B}{5}_{16}$. Octal — group into 3s from the right, padding the leftmost group: $10110101\to \underset{2}{\underbrace{010}}\underset{6}{\underbrace{110}}\underset{5}{\underbrace{101}}$. So $265_8$. Sanity via decimal: $10110101_2=128+32+16+4+1=181_{10}$; and ${\text{B5}}_{16}=11\cdot 16+5=181$ ✓, $265_8=2\cdot 64+6\cdot 8+5=128+48+5=181$ ✓.

Here $101101_2$ has exactly $6$ bits, a clean multiple of $3$, so left and right grouping give the same split $101101$ → $55_8$. The answer $55_8$ is actually correct ($55_8=5\cdot 8+5=45$; and $101101_2=32+8+4+1=45$ ✓). The lesson: left-grouping is silently equivalent only when the bit-count is already a multiple of the group size. The trap bites the moment the count isn't a multiple of 3 (or 4) — then the padding must go on the left, which is impossible if you started chopping there. So "it worked once" is a false friend.

Why not direct: hex groups are 4 bits, octal groups are 3 bits; $4$ and $3$ share no common alignment, so there is no digit-by-digit hex↔octal map. Binary is the common middle-man (both align to it). Expand each hex digit to 4 bits: $\text{A}=1010$, $7=0111$ → $10100111_2$. Regroup into 3s from the right: $10100111\to \underset{2}{\underbrace{010}}\underset{4}{\underbrace{100}}\underset{7}{\underbrace{111}}$. So ${\text{A7}}_{16}=247_8$. Check: ${\text{A7}}_{16}=10\cdot 16+7=167$; $247_8=2\cdot 64+4\cdot 8+7=128+32+7=167$ ✓.

${\text{FF}}_{16}$ = two hex digits = $8$ bits, all ones ($F=1111$), so $11111111_2$. $377_8$: $3=011,7=111,7=111$ → $011111111=011111111_2$, which is also $8$ ones (leading zero drops). Both are $11111111_2$, so they are equal — both are $255_{10}$.

Step — why $256+n$: in 8-bit two's complement, the negative number $n$ is stored as the unsigned pattern $256+n$ (because $2^8=256$ wraps around). Here $256+(-6)=250$. Convert $250$ to binary: $250=128+64+32+16+8+2=11111010_2$. Group into 4s from the right: $\underset{\text{F}}{\underbrace{1111}}\underset{\text{A}}{\underbrace{1010}}$ → $\mathbf{F}{\mathbf{A}}_{16}$. So $-6$ in an 8-bit register is $11111010_2={\text{FA}}_{16}$. Check: ${\text{FA}}_{16}=15\cdot 16+10=250$, and $250-256=-6$ ✓.

"Keep low 4, clear top 4" means the mask has $1$s in positions $0$–$3$ and $0$s in $4$–$7$: $00001111_2$. Group into 4s: $\underset{0}{\underbrace{0000}}\underset{\text{F}}{\underbrace{1111}}$ → $0{\mathbf{F}}_{16}$ (often written 0x0F). Why hex is obvious: each hex digit is one nibble. A digit of $\text{F}$ = "all four bits on", a digit of $0$ = "all four off". So a nibble-aligned mask reads straight off the hex: 0x0F literally spells "off-nibble, on-nibble."

Three hex digits × 4 bits = up to $12$ bits. Expand: $1=0001,\text{F}=1111,4=0100$ → $000111110100_2$. Drop leading zeros for the minimum bit count: $111110100_2$ = $9$ significant bits (the top bit is at position 8, so bits $0$–$8$, i.e. 9 bits minimum). Octal — regroup the full 12-bit string into 3s from the right: $000111110100\to \underset{0}{\underbrace{0}}\underset{111}{\underbrace{7}}\underset{110}{\underbrace{6}}\underset{100}{\underbrace{4}}$ = $0764_8$, i.e. $764_8$. Check: $\text{0x1F4}=1\cdot 256+15\cdot 16+4=256+240+4=500$; $764_8=7\cdot 64+6\cdot 8+4=448+48+4=500$ ✓.

Derive the group size: $4=2^2$ so one base-4 digit = 2 bits; $16=2^4$ so one hex digit = 4 bits. Thus 2 base-4 digits = 1 hex digit (because $2+2=4$ bits). Groups align because both are powers of $2$ — the same "middle-man binary" idea, now with the friendlier fact that $4$ is itself a power of two. Expand each base-4 digit to 2 bits ($0=00,1=01,2=10,3=11$): $3121\to 11011001=11011001_2$. Group into 4s from the right: $\underset{\text{D}}{\underbrace{1101}}\underset{9}{\underbrace{1001}}$ → $\mathbf{D}{9}_{16}$. Check: $3121_4=3\cdot 64+1\cdot 16+2\cdot 4+1=192+16+8+1=217$; ${\text{D9}}_{16}=13\cdot 16+9=217$ ✓.

${\text{5C}}_{16}$: $5=0101,\text{C}=1100$ → $01011100_2$. Rotate left by 2: take the top 2 bits 01 off the left and append them on the right. The middle bits 011100 shift up. $01011100\to (011100)(01)=01110001_2$. Regroup: $\underset{7}{\underbrace{0111}}\underset{1}{\underbrace{0001}}$ → $71_{16}$. Check by value: $01011100_2=92$. Rotate-left-2 of an 8-bit value $=((92\ll 2)\text{mod}256)+(92\gg 6)=(368\mathrm{mod}256)+1=112+1=113=01110001_2={\text{71}}_{16}$ ✓.

A power of two $2^k$ in binary is a single $1$ followed by $k$ zeros. When we group into 4s from the right, that lone $1$ lands at bit position $k$, i.e. in group number $\lfloor k/4\rfloor$, at offset $k\mathrm{mod}4$ inside it.

• If $k\mathrm{mod}4=0$, the $1$ sits at the bottom of a group → that group is $0001_2=1$, and it's a single nonzero hex digit ($1$). Example: $2^8=\text{0x100}$, leading digit $1$.
• If $k\mathrm{mod}4\ne 0$, the $1$ sits higher inside the group, giving that group the value $2^{(k\mathrm{mod}4)}$ ($=2,4,\text{or }8$) — still one nonzero hex digit, e.g. $2^{10}$. Let's verify $2^{10}=1024$: binary is $1$ then ten $0$s = $10000000000_2$ (11 bits). Group into 4s from the right: $\underbrace{100}\underbrace{0000}\underbrace{0000}\to$ pad left → $\underset{4}{\underbrace{0100}}\underset{0}{\underbrace{0000}}\underset{0}{\underbrace{0000}}$ = $400_{16}$. Here $k=10$, $10\mathrm{mod}4=2$, so the top hex digit is $2^2=4$ — a single nonzero hex digit ($4$), just not the digit $1$. The claim as stated ("digit $1$ only when $k$ is a multiple of 4") holds: only $k\equiv 0(\mathrm{mod}4)$ gives leading digit exactly 1.

L1.1 C=12, E=14 · L1.2 "10"=b · L1.3 legal $47_8,17_8$ · L2.1 $58$ · L2.2 $180$ · L2.3 ${\text{B4}}_{16}$ · L2.4 ${\text{B5}}_{16},265_8$ · L3.1 $55_8$ (equal by luck) · L3.2 $247_8$ · L3.3 equal ($255$) · L4.1 ${\text{FA}}_{16}$ · L4.2 0x0F · L4.3 9 bits, $764_8$ · L5.1 ${\text{D9}}_{16}$ · L5.2 $71_{16}$ · L5.3 $\text{0x400}$.