3-input AND requirement: For a three-input AND gate, what condition on the inputs guarantees a LOW output?

Introduction / Context:
Understanding how multi-input AND gates behave is critical when designing enable signals, chip selects, and safety interlocks. The AND function outputs HIGH only if all inputs are HIGH, so identifying the condition for a LOW output follows directly from its truth table.

Given Data / Assumptions:

• A 3-input AND gate with inputs A, B, C and output X = A·B·C.
• Binary logic levels are 0 and 1.
• No inversion or special active-LOW semantics are present on inputs/outputs.

Concept / Approach:
For AND, the product is 1 only when every factor is 1. Therefore, the presence of even a single 0 anywhere in the inputs forces the product to 0, guaranteeing a LOW output.

Step-by-Step Reasoning:
Write the output: X = A·B·C.If A = 0, then X = 0·B·C = 0 regardless of B and C.Similarly for B = 0 or C = 0, the result is 0.Thus, “at least one input LOW” assures X is LOW.

Verification / Alternative check:
Truth table confirms that only the row A=1, B=1, C=1 yields X=1; all other seven combinations contain at least one 0 and produce X=0.

Why Other Options Are Wrong:

• “All LOW” is sufficient but not necessary; a single LOW already guarantees LOW output.
• “All HIGH” yields HIGH, not LOW.
• “At least one HIGH” is insufficient, since two other inputs could be LOW.

Common Pitfalls:

• Confusing AND with OR, where the presence of at least one HIGH is significant.
• Assuming symmetry with NAND; NAND outputs the complement of AND and changes the required conditions.

Final Answer:
At least one input must be LOW.