Binary multiplication practice When multiplying the decimal numbers 13 × 11 using binary arithmetic (13 = 1101₂ and 11 = 1011₂), what is the third partial product generated during the shift-and-add process?

Introduction / Context:
This problem checks your understanding of binary multiplication by hand, specifically identifying the intermediate (partial) products that arise when a multiplicand is conditionally shifted and added based on the multiplier bits.

Given Data / Assumptions:

• Multiplicand: 13₁₀ = 1101₂.
• Multiplier: 11₁₀ = 1011₂ (bits processed from least significant to most significant).
• Standard shift-and-add binary multiplication procedure.

Concept / Approach:
For each bit of the multiplier (from LSB to MSB), write a partial product: either the multiplicand left-shifted by the bit position (if that bit is 1) or a row of zeros (if that bit is 0). The “third partial product” corresponds to the third processed bit of the multiplier (bit index 2, counting from 0 at the LSB).

Step-by-Step Solution:
Write numbers: multiplicand = 1101₂, multiplier = 1011₂.Bit 0 (LSB) of multiplier = 1 → 1st partial product = 1101.Bit 1 of multiplier = 1 → 2nd partial product = 11010 (one left shift).Bit 2 of multiplier = 0 → 3rd partial product = 0000 (all zeros, two-position row).Bit 3 of multiplier = 1 → 4th partial product = 1101000 (three left shifts).

Verification / Alternative check:
If the third multiplier bit is zero, no contribution occurs at that stage, so the corresponding row must be zeros, confirming the choice.

Why Other Options Are Wrong:
100000 / 100001 / 1011: These represent nonzero rows; they would imply a 1 in the third multiplier bit, which is not the case.

Common Pitfalls:
Counting partial products from the MSB side or forgetting that a 0 multiplier bit yields an all-zero partial product.

Final Answer:
0000