--- title: "Restoring division of 10 by 3" subject: "Computer Organization and Architecture" chapter: 5 key: "semester_5/computer_organization_and_architecture:restoring-10-3" source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/restoring-10-3 site: BCT Engineering --- # Restoring division of 10 by 3 Computer Organization and Architecture · Chapter 5: Computer Arithmetic - Asked in: 2075 Chaitra, 7 marks - Video: [Restoring Division Algorithm: Flowchart, Circuit & Example Explained | COA](https://www.youtube.com/watch?v=Ek-Ea2NnjeQ) — Engineering Funda - 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/restoring-10-3 - Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5 ## Question Perform 10/3 using the restoring division algorithm. ## Short answer Restoring division shifts A,Q left, subtracts divisor, and restores A if negative while setting quotient bits. For 10 divided by 3, the quotient is 3 (0011) and the remainder is 1 (00001). ## Answer Restoring division: shift left A,Q; A = A − M; if A < 0 restore and Q0 = 0, else Q0 = 1; repeat n times. Dividend = 10 = 1010 (Q), divisor M = 3 = 00011 (n+1 = 5 bits), A = 0, n = 4. ```text Step Action A Q - Initial 00000 1010 1 Shift left A,Q 00001 0100 A = A - M (1 - 3 = -2) 11110 0100 A<0: Q0=0, restore A 00001 0100 2 Shift left A,Q 00010 1000 A = A - M (2 - 3 = -1) 11111 1000 A<0: Q0=0, restore A 00010 1000 3 Shift left A,Q 00101 0000 A = A - M (5 - 3 = 2) 00010 0000 A>=0: Q0=1 00010 0001 4 Shift left A,Q 00100 0010 A = A - M (4 - 3 = 1) 00001 0010 A>=0: Q0=1 00001 0011 ``` **Answer: 10 ÷ 3: quotient = 3 (0011), remainder = 1 (00001)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is M = 00011 (5 bits)? A. A and M are n+1 bits, n = 4 B. Divisor needs 5 bits C. To hold quotient D. Because 10 is 5 bits Answer: A. Dividend is 4-bit so the adder width is n+1 = 5 bits. (Key point: Divisor width) 2. In step 1, A = 1 - 3 = -2. What is done? A. Keep A and Q0 = 1 B. Restore A and Q0 = 0 C. Keep A and Q0 = 0 D. Restore A and Q0 = 1 Answer: B. Negative result: restore and quotient bit 0. (Key point: Step 1) 3. What are quotient and remainder of 10 / 3? A. Quotient 3, remainder 1 B. Quotient 2, remainder 4 C. Quotient 1, remainder 3 D. Quotient 3, remainder 0 Answer: A. Q = 0011 and A = 00001; 3 x 3 + 1 = 10. (Key point: Result) ## Related in Computer Organization and Architecture - [Restoring division algorithm and hardware](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Arestoring-division) — Computer Organization and Architecture, chapter 5, asked 2× - [Non-restoring division of 10 by 5](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Anonrestoring-10-5) — Computer Organization and Architecture, chapter 5 - [Non-restoring division with flowchart](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Anonrestoring-explain) — Computer Organization and Architecture, chapter 5, asked 3× - [Signed division and non-restoring division](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Anonrestoring-division) — Computer Organization and Architecture, chapter 5 - [Restoring division of 31 by 13](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Arestoring-31-13) — Computer Organization and Architecture, chapter 5 - [Restoring vs non-restoring division](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_organization_and_architecture%3Arestoring-vs-nonrestoring) — Computer Organization and Architecture, chapter 5, asked 5×