For a multi-input NOR gate, what must the input conditions be to obtain a logic-high output at the gate's output?

Introduction / Context:
The NOR gate is the negation of the OR function. Understanding its truth table is vital because NOR (like NAND) is functionally complete—any Boolean function can be built from it. Here we recall the specific input condition that produces an asserted (high) output from a NOR gate.

Given Data / Assumptions:

• Positive logic convention: high = 1, low = 0.
• NOR = NOT(OR of inputs).
• The device may have two or more inputs.

Concept / Approach:

OR outputs high if any input is high. NOR inverts that outcome, thus it outputs high only in the single case where the OR would be low—namely when all inputs are low. Therefore, every input must be low to obtain a high output from NOR.

Step-by-Step Solution:

Verification / Alternative check:

Truth tables for 2-, 3-, and 4-input NOR devices consistently show a single high-output row corresponding to all-zero inputs.

Why Other Options Are Wrong:

'high' or 'some low some high' would cause the OR to be 1, making NOR = 0. '1's' is just another way to say high inputs, which likewise drive NOR low.

Common Pitfalls:

Mixing up NOR with NAND (which outputs low only if all inputs are high) and forgetting that NOR’s asserted output corresponds to the all-zero input combination.

Final Answer:

low