When multiplying 13 × 11 in binary (13 = 1101, 11 = 1011), what is the third partial product generated during the standard shift-and-add multiplication process (counting from the least significant multiplier bit upward)?

Introduction / Context:Binary multiplication by hand mirrors decimal long multiplication: for each '1' in the multiplier, you write a shifted copy of the multiplicand and then sum the partial products. This question asks specifically for the third partial product when multiplying 13 by 11 in binary.

Given Data / Assumptions:

• Multiplicand: 13 = 1101.
• Multiplier: 11 = 1011 (bits b0=1, b1=1, b2=0, b3=1 from LSB to MSB).
• Partial products are ordered from LSB upward.

Concept / Approach:For each multiplier bit b, the partial product equals (multiplicand << i) if b = 1; otherwise it is all zeros for that position. The 'third partial product' corresponds to i = 2 (the third bit from LSB, b2).

Step-by-Step Solution:1) b0 = 1 → P0 = 1101.2) b1 = 1 → P1 = 1101 shifted left 1 = 11010.3) b2 = 0 → P2 = 00000000 (all zeros for that row).4) b3 = 1 → P3 = 1101 shifted left 3 = 1101000.

Verification / Alternative check:Compute the final product to validate the steps: 13 × 11 = 143 decimal. Binary sum of partials P0 + P1 + P2 + P3 equals 10001111, which is indeed 143. Hence P2 is zero as expected.

Why Other Options Are Wrong:1011: that is the unshifted multiplier or a possible P0 for a different setup, not the third partial product here.100000 or 100001: look like shifted patterns but do not correspond to b2 = 0.00110100: unrelated shifted value for this step.

Common Pitfalls:Counting partial products from the MSB instead of the LSB, or confusing multiplicand/multiplier roles. Always index partials by multiplier bit position from least significant upward in standard long multiplication.

Final Answer:00000000