Digital logic — 3-input AND gate behavior For an AND gate with three inputs A, B, and C, identify the exact input condition under which the output is HIGH (logic 1).

Introduction / Context:This question checks fundamental understanding of the AND operation in digital electronics. AND gates are ubiquitous in control logic, enable lines, and safety interlocks, where a HIGH output should occur only when all required conditions are simultaneously true.

Given Data / Assumptions:

• A standard 3-input AND gate with inputs A, B, and C.
• Logic levels: 0 represents LOW, 1 represents HIGH.
• No inversion bubbles are present on inputs or output.

Concept / Approach:The Boolean equation for a 3-input AND gate is X = A * B * C (often written X = ABC). The product of binary variables equals 1 only if every factor equals 1. If any input is 0, the product becomes 0 immediately, forcing a LOW output.

Step-by-Step Solution:Let X = A * B * C.Check case A = 1, B = 1, C = 1: X = 1 * 1 * 1 = 1.If any one input is 0 (e.g., A = 1, B = 1, C = 0), then X = 1 * 1 * 0 = 0.Therefore, only the all-ones input produces a HIGH output.

Verification / Alternative check:Construct the 3-input AND truth table. Out of the 8 possible input triplets, exactly one row (1,1,1) yields X = 1; the other seven rows contain at least one zero and yield X = 0. This matches the product rule of Boolean algebra.

Why Other Options Are Wrong:

• A = 1, B = 1, C = 0 → X = 0 because one input is LOW.
• A = 0, B = 0, C = 0 → X = 0 since all are LOW.
• A = 1, B = 0, C = 1 → X = 0 because B is LOW.

Common Pitfalls:

• Confusing AND with OR. OR would output HIGH if any input is HIGH; AND requires all inputs HIGH.
• Assuming two HIGHs are “enough”—for AND, every input must be HIGH without exception.

Final Answer:A = 1, B = 1, C = 1