Priority encoder 74148 (octal-to-binary, active-low): When input I6 is the active request, what is the 3-bit data output (Y2 Y1 Y0)? Assume the device follows standard 74148 conventions: inputs active LOW, outputs active LOW, and I7 has the highest priority.

Introduction / Context:
The 74148 is an MSI priority encoder that accepts eight requests on active-low inputs (I0–I7) and outputs a 3-bit code on active-low outputs (Y2 Y1 Y0). It asserts the code corresponding to the highest-priority active input (I7 highest). Because both inputs and outputs are active low, learners must mentally invert logic levels to read the resulting binary code correctly. This question tests that mapping for I6 being active.

Given Data / Assumptions:

• Inputs I0–I7 are active LOW; “active” means driven to 0.
• Outputs Y2 Y1 Y0 are also active LOW (low-true).
• Only I6 is active; all higher-priority inputs (I7) are inactive (HIGH).
• Enable/GS/EO pins are in their enabling states so the code is presented.

Concept / Approach:
For a given active input In, the true binary code is simply n expressed in 3 bits. However, the 74148 provides low-true outputs, so the pins Y2 Y1 Y0 equal the bitwise inversion of that 3-bit code. Example: for n = 7 (111 true), Y outputs are 000. Applying the same inversion for n = 6 (110 true) yields Y = 001.

Step-by-Step Solution:

Verification / Alternative check:
Check limiting cases: For I7 active, true code is 111 and Y = 000; for I0 active, true code is 000 and Y = 111. The sequence matches the expected monotone pattern, validating the inversion rule used for I6 → 001.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting that both inputs and outputs are active low, or assuming a straight binary output without inversion, which leads to an off-by-inversion error.

Final Answer:
001