Binary width awareness: How many binary bits are required to represent a single hexadecimal digit (0–F)?

Introduction / Context:
Hexadecimal is a compact notation for binary values. Because hex is base 16 and 16 equals 2^4, there is a direct one-to-one mapping between a single hex digit and four binary bits, known as a nibble.

Given Data / Assumptions:

• Hex digit range: 0–F (0–15 decimal).
• Binary digits (bits) are base-2.
• Goal: minimal bit width to encode any hex digit.

Concept / Approach:
The number of bits required to represent N distinct values is the smallest k such that 2^k ≥ N. For a hex digit, N = 16, and 2^4 = 16, so k = 4. Therefore each hex digit corresponds exactly to a 4-bit binary group, enabling easy grouping and conversion between binary and hex.

Step-by-Step Solution:
Set N = 16.Find k with 2^k ≥ 16 → k = 4.Therefore, one hex digit requires 4 bits.

Verification / Alternative check:
Examples: A₁₆ = 10₁₀ = 1010₂ (4 bits), F₁₆ = 15₁₀ = 1111₂ (4 bits). The mapping is exact.

Why Other Options Are Wrong:
3 bits hold only 8 values, 6 bits hold 64, and 8 bits hold 256—over/under the needed capacity for a single hex digit.

Common Pitfalls:
Using 8 bits per hex digit out of habit from byte-aligned thinking; remember, 1 hex digit = 4 bits, 2 hex digits = 8 bits (one byte).

Final Answer:
4