Two’s-complement subtraction — 43 − 15 using 6-bit representation When subtracting decimal 15 from 43 in binary using two’s complement (6-bit width), which two’s-complement value must be added to 43?

Introduction / Context:
Binary subtraction is commonly performed by adding the two’s-complement of the subtrahend. This problem tests whether you can form the correct two’s-complement for a given decimal number using a fixed bit width and apply it to a subtraction example.

Given Data / Assumptions:

• Compute 43 − 15 using 6 bits.
• Two’s-complement formation: invert bits (one’s-complement) then add 1.
• 15 in 6-bit binary is 00 1111.

Concept / Approach:

To subtract B from A, compute A + (two’s-complement of B). The two’s-complement of B is formed by inverting all bits of B (with respect to the chosen width) and adding 1. Ensure the correct width to avoid sign or overflow mistakes.

Step-by-Step Solution:

Verification / Alternative check:

Decimal check: 43 − 15 = 28. Binary result 011100 is 28, confirming that 110001 was the correct two’s-complement to add.

Why Other Options Are Wrong:

• 110000: this is only the one’s-complement; you must add 1.
• 101011 and 011100: these are 43 and 28 respectively, not the required complement.
• 001111: original +15 rather than its negative.

Common Pitfalls:

• Forgetting to fix the bit width before forming complements.
• Failing to add 1 after inversion.

Final Answer:

110001