--- title: "Booth multiplication of 6 × 7" subject: "Computer Organization and Architecture" chapter: 5 key: "semester_5/computer_organization_and_architecture:booth-6x7" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/booth-6x7 site: BCT Engineering --- # Booth multiplication of 6 × 7 Computer Organization and Architecture · Chapter 5: Computer Arithmetic - Asked in: 2067 Asar (old course), 4 marks - Video: [Booth's Algorithm With Example | booths | booths algo](https://www.youtube.com/watch?v=cWfaw7b3jKY) — Always Learn More - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/booth-6x7 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5 ## Question Multiply 6 × 7 using the Booth algorithm. ## Short answer Using 4-bit Booth's algorithm, 6 (0110) × 7 (0111), -M=1010. The run of 1s requires one subtraction and one addition, giving final A,Q=0010 1010 = 42. ## Answer Booth's algorithm with 4-bit registers (6 = 0110, 7 = 0111). A run of 1s in the multiplier (0111) needs one subtraction at the start and one addition after the run. Multiplicand M = 6 = 0110, multiplier Q = 7 = 0111, -M = 1010, n = 4 bits. ```text Step Action A Q Q-1 - Initial 0000 0111 0 1 Q0Q-1=10: A = A - M 1010 0111 0 ASR 1101 0011 1 2 Q0Q-1=11: no change 1101 0011 1 ASR 1110 1001 1 3 Q0Q-1=11: no change 1110 1001 1 ASR 1111 0100 1 4 Q0Q-1=01: A = A + M 0101 0100 1 ASR 0010 1010 0 ``` Result in A,Q = 0010 1010 = 42. **Answer: 6 × 7 = 42** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Multiplier 0111 has a run of 1s. What does Booth do? A. Subtract at the start of the run and add after it B. Add three times C. Subtract three times D. Add at the start and subtract after Answer: A. 7 = 8 - 1: one subtract and one add. (Key point: Run of 1s) 2. For M = 6 = 0110, -M is A. 1001 B. 1010 C. 0110 D. 1011 Answer: B. 2's complement of 0110 is 1010. (Key point: -M value) 3. 6 x 7 = ? A. 13 = 0000 1101 B. -42 = 1101 0110 C. 42 = 0010 1010 D. 42 = 0101 0100 Answer: C. A:Q = 0010 1010 = 42. (Key point: Result) ## Related in Computer Organization and Architecture - [Booth multiplication algorithm and hardware](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-algorithm) — Computer Organization and Architecture, chapter 5, asked 8× - [Booth multiplication of -5 × -9](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-5x-9) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of 5 × -6](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-5x-6) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of -7 × -10](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-7x-10) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of 10 × (-5)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-10x-5) — Computer Organization and Architecture, chapter 5, asked 5× - [Booth multiplication of -6 × -11](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-6x-11) — Computer Organization and Architecture, chapter 5, asked 5×