Hex-to-binary identity — validate 3C1D₁₆: Confirm or refute: “3C1D₁₆ equals 1111 0000 0111 01₂” (i.e., 11110000011101₂ when spaces are removed).

Introduction / Context:Translating hex to binary is a matter of replacing each hex digit with its 4-bit binary equivalent. This question tests that direct mapping and your comfort with optional leading zeros in binary representations.

Given Data / Assumptions:

• Hex numeral: 3C1D₁₆.
• Hex digit map: 0–9, A=10, B=11, C=12, D=13, E=14, F=15.
• Binary is allowed to omit leading zeros unless a fixed width is required.

Concept / Approach:Map each hex digit to 4 bits and concatenate. 3→0011, C→1100, 1→0001, D→1101. Concatenate to get 0011 1100 0001 1101. Dropping the leading two zeros yields 11 1100 0001 1101, which is 11110000011101₂. Therefore the given binary matches the hex value, acknowledging optional omission of leading zeros.

Step-by-Step Solution:3 → 0011C → 11001 → 0001D → 1101Concatenate: 0011 1100 0001 1101 → remove leading zeros → 11110000011101₂.

Verification / Alternative check:Group 11110000011101₂ into nibbles from the right: 11 1100 0001 1101 → pad left with two zeros → 0011 1100 0001 1101 → 3C1D₁₆.

Why Other Options Are Wrong:Insisting on exactly 16 bits is unnecessary unless a fixed-width field is specified; correctness is about value, not width. The mapping remains valid with or without leading zeros.

Common Pitfalls:Forgetting to pad the most significant nibble when regrouping; misreading C (1100) and D (1101) patterns.

Final Answer:Correct