8-input XOR output evaluation An 8-input XOR gate produces Y = 1 when an odd number of its inputs are at logic 1. For the input vector ordered A to H, which of the following bit patterns will make Y = 1?

Introduction / Context:
An exclusive-OR (XOR) gate outputs logic 1 if and only if the number of 1s at its inputs is odd. Multi-input XORs are common in parity generators/checkers and error-detection coding. Recognizing the odd-parity rule enables quick mental evaluation without writing full truth tables.

Given Data / Assumptions:

• Gate: 8-input XOR.
• Rule: output Y = 1 when count_of_ones is odd; otherwise Y = 0.
• Input ordering is A through H; options are 8-bit strings.

Concept / Approach:
Count the number of 1s in each candidate bit pattern. Odd count implies Y = 1; even count implies Y = 0. This is equivalent to computing parity of the vector.

Step-by-Step Solution:

Verification / Alternative check:
Pairwise XOR accumulation confirms the same results: XOR of an odd number of 1s yields 1; XOR of an even number of 1s yields 0. Using parity intuition gives identical outcomes.

Why Other Options Are Wrong:

• B, C, D each contain an even number of 1s, so an 8-input XOR outputs 0.

Common Pitfalls:
Miscounting the ones; confusing XOR with XNOR (which outputs 1 for an even number of 1s). Always confirm the odd vs. even parity rule for XOR vs. XNOR respectively.

Final Answer:
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