Operational condition for NAND: For a multi-input NAND gate, under which input condition will the output be 1 (HIGH)?

Introduction / Context:
NAND is ubiquitous because it is functionally complete. Knowing when its output is HIGH clarifies many design shortcuts and fault predictions.

Given Data / Assumptions:

• NAND = NOT(AND).
• Inputs can be any pattern of 0 and 1.

Concept / Approach:
AND outputs 1 only when all inputs are 1. Therefore, NAND outputs 0 only when all inputs are 1; in every other case (i.e., when any input is 0), NAND outputs 1.

Step-by-Step Solution:
1) Evaluate AND for the input pattern.2) Invert to obtain NAND.3) Conclude output 1 occurs whenever at least one input is 0.

Verification / Alternative check:
Construct the two-input NAND table and observe the three rows producing 1 versus the single all-1 row producing 0.

Why Other Options Are Wrong:
“Only when all inputs = 0” confuses NOR with NAND; “Whenever a 1…” is AND-centric; “Only when all = 1” is exactly the LOW case for NAND.

Common Pitfalls:
Mixing NOR and NAND rules; forgetting the inversion step relative to AND.

Final Answer:
Whenever a 0 is present at an input