Introduction:
This problem assesses binary multiplication fluency. We multiply two unsigned 4-bit numbers, 1001₂ and 1100₂, and report the 8-bit product. Accurate partial-product addition and place shifting are the keys.
Given Data / Assumptions:
- Multiplicand = 1001₂ (decimal 9).
- Multiplier = 1100₂ (decimal 12).
- Unsigned binary arithmetic; report an 8-bit result.
Concept / Approach:
Multiply as in decimal, but in base 2. Each 1 bit in the multiplier generates a shifted copy of the multiplicand; sum these partial products. Since 9 × 12 = 108 in decimal, the binary should equal the 8-bit representation of 108.
Step-by-Step Solution:
Step 1: 1001₂ = 9, 1100₂ = 12.Step 2: 9 × 12 = 108 (decimal) for a quick cross-check.Step 3: Convert 108 to binary: 108 = 64 + 32 + 8 + 4 = 2^6 + 2^5 + 2^3 + 2^2 ⇒ 01101100₂.Step 4: Verify via partial products: 1001 × 1100 = (1001 × 1100) = (1001 << 2) + (1001 << 3) = 00100100 + 01001000 = 01101100.
Verification / Alternative check:
Decimal-binary round-trip confirms: 01101100₂ = 64+32+8+4 = 108, matching 9 × 12.
Why Other Options Are Wrong:
01110001 = 113 decimal (too high).01111000 = 120 decimal (too high).01101110 = 110 decimal (close but not exact).
Common Pitfalls:
Misplacing shifts (left by bit position of each 1 in the multiplier).Arithmetic carry errors when adding partial products.
Final Answer:
01101100
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