Use the place value table. Write out the binary number and multiply each digit by its place value, then add the results.
10110101₂ = 128 + 32 + 16 + 4 + 1 = 181₁₀
Work from the largest place value down. Ask: "Is 128 ≤ my number?" If yes, put a 1 and subtract it.
128 ≤ 45? No → 0
64 ≤ 45? No → 0
32 ≤ 45? Yes → 1, remainder = 45 − 32 = 13
16 ≤ 13? No → 0
8 ≤ 13? Yes → 1, remainder = 13 − 8 = 5
4 ≤ 5? Yes → 1, remainder = 5 − 4 = 1
2 ≤ 1? No → 0
1 ≤ 1? Yes → 1, remainder = 0
45₁₀ = 00101101₂
Split the binary number into groups of 4 (nibbles), working from the right. Convert each nibble to its hex digit.
Split: 1011 | 0111
1011 = 8+2+1 = 11 = B
0111 = 4+2+1 = 7 = 7
10110111₂ = B7₁₆
Expand each hex digit into exactly 4 binary bits.
3 = 0011
F = 15 = 1111
3F₁₆ = 00111111₂
Use place values for base 16: 16¹ = 16, 16⁰ = 1. For three digits: 16² = 256.
(2 × 16) + (A × 1) = (2 × 16) + (10 × 1) = 32 + 10 = 42₁₀
(1 × 256) + (F × 16) + (4 × 1) = 256 + 240 + 4 = 500₁₀
Repeatedly divide by 16. The remainders (read bottom to top) give the hex digits.
200 ÷ 16 = 12 remainder 8 → digit: 8
12 ÷ 16 = 0 remainder 12 → digit: C
Read remainders bottom to top → C8₁₆
Check: (12 × 16) + 8 = 192 + 8 = 200 ✓
| Decimal | Binary (4-bit) | Hex |
|---|---|---|
| 0 | 0000 | 0 |
| 5 | 0101 | 5 |
| 10 | 1010 | A |
| 12 | 1100 | C |
| 15 | 1111 | F |
| 16 | 0001 0000 | 10 |
| 255 | 1111 1111 | FF |
8 questions · 18 marks
| Method | How to do it |
|---|
Timed exam conditions.