Determine the minimum number of bits required to uniquely encode 26 letters, 10 symbols, and 10 numerals (total distinct characters) using a fixed-length binary code.

Introduction / Context:Choosing code length is a basic information theory and coding design task. For a fixed-length binary code, the number of representable symbols is 2^n for n bits. We must cover all distinct characters with the smallest n.

Given Data / Assumptions:

• Alphabet size: 26 letters + 10 symbols + 10 numerals = 46 characters.
• Codewords are fixed-length and binary.
• No parity or extra control bits are required for this calculation.

Concept / Approach:Find the smallest integer n such that 2^n ≥ 46. Compute 2^5 = 32 and 2^6 = 64; therefore n = 6 satisfies the requirement, while n = 5 does not.

Step-by-Step Solution:

Verification / Alternative check:Even if some codes remain unused (64 − 46 = 18), 6 bits is still minimal; 5 bits cannot encode all 46 characters.

Why Other Options Are Wrong:

• 5: represents only 32 distinct values, insufficient.
• 2 or 3: far too small (4 or 8 values).
• None of the above: incorrect because 6 is correct.

Common Pitfalls:Forgetting to include symbols, confusing variable-length with fixed-length codes, or assuming exact power-of-two sizes are mandatory.

Final Answer:6