Identify the logic gate whose output is 1 if and only if at least one input is 0 (i.e., it outputs 0 only when all inputs are 1).

Introduction / Context:
Understanding the behavior of basic logic gates is essential for designing and simplifying digital circuits. This question asks for the gate that produces a high output in every case except when all inputs are high. Recognizing this pattern allows quick identification without building full truth tables every time.

Given Data / Assumptions:

• The gate is multi-input (2 or more) but the rule must hold for any number of inputs.
• Output is 0 only when all inputs are 1.
• In every other input combination, the output is 1.

Concept / Approach:
The NAND function is the logical NOT of AND. Because AND outputs 1 only when all inputs are 1, NAND outputs 0 only in that same case and outputs 1 otherwise. This matches the statement “output is 1 if at least one input is 0.” NAND is functionally complete, meaning any Boolean function can be built using only NAND gates, which is why it is so common in IC design.

Step-by-Step Solution:
Recall AND truth: all 1s → 1, otherwise → 0.Invert (NAND): all 1s → 0, otherwise → 1.Compare with the condition: “output 1 if at least one input is 0.”Therefore, the gate is NAND.

Verification / Alternative check:
For two inputs, list cases: 00→1, 01→1, 10→1, 11→0 for NAND. This exactly matches the described behavior.

Why Other Options Are Wrong:
AND outputs 0 whenever any input is 0; OR outputs 1 if any input is 1 (different rule); NOR outputs 1 only when all inputs are 0; “None” is false since NAND is correct.

Common Pitfalls:
Confusing “at least one input is 0” (NAND) with “at least one input is 1” (OR). Also mixing up NOR (only 1 when all 0s) with NAND (only 0 when all 1s).

Final Answer:
NAND gate