--- title: "Booth multiplication 7×3" subject: "Computer Organization and Architecture" chapter: 5 key: "semester_5/computer_organization_and_architecture:booth-7x3" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/booth-7x3 site: BCT Engineering --- # Booth multiplication 7×3 Computer Organization and Architecture · Chapter 5: Computer Arithmetic - Asked in: 2074 Chaitra, 6 marks - Video: [Multiply 7*3 using booth's algorithm | Computer Architecture BE Computer 4th Sem](https://www.youtube.com/watch?v=CheGgcrPxco) — BE Computer - 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-7x3 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5 ## Question Perform multiplication 7×3 using Booth algorithm. ## Short answer Booth's algorithm computes 7×3. M=0111, Q=0011, -M=1001, n=4. After 4 steps of add/subtract and arithmetic shift right, A,Q = 0001 0101 = 21. So 7 × 3 = 21. ## Answer Booth's algorithm: look at (Q0, Q-1); 10 means A = A − M, 01 means A = A + M, 00/11 mean no operation; then arithmetic shift right (ASR) of A, Q, Q-1. Repeat n times. Multiplicand M = 7 = 0111, multiplier Q = 3 = 0011, -M = 1001, n = 4 bits. ```text Step Action A Q Q-1 - Initial 0000 0011 0 1 Q0Q-1=10: A = A - M 1001 0011 0 ASR 1100 1001 1 2 Q0Q-1=11: no change 1100 1001 1 ASR 1110 0100 1 3 Q0Q-1=01: A = A + M 0101 0100 1 ASR 0010 1010 0 4 Q0Q-1=00: no change 0010 1010 0 ASR 0001 0101 0 ``` Result in A,Q = 0001 0101 = 21. **Answer: 7 × 3 = 21** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are M, -M and Q in the 4-bit example 7 x 3? A. M = 0111, -M = 1001, Q = 0011 B. M = 0011, -M = 1101, Q = 0111 C. M = 0111, -M = 1000, Q = 0011 D. M = 1001, -M = 0111, Q = 0011 Answer: A. -7 in 4-bit 2's complement is 1001. (Key point: Setup) 2. At step 1 Q0Q-1 = 10, so: A. A = A - M = 1001 then ASR B. A = A + M C. no operation D. shift left Answer: A. The start of a run of 1s triggers subtraction. (Key point: Step 1) 3. Step 3 has Q0Q-1 = 01 which means: A. A = A + M B. A = A - M C. no operation D. stop Answer: A. The end of the run adds M. (Key point: Step 3) 4. Final A:Q is: A. 0001 0101 = 21 B. 0010 1010 = 42 C. 1110 1011 = -21 D. 0000 0111 = 7 Answer: A. 7 x 3 = 21 = 00010101. (Key point: Result) ## Close variants set in other exams - [Booth algorithm and 8 × 4](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Abooth-8x4) — Computer Organization and Architecture, chapter 5, asked 2× ## 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×