---
title: "Non-restoring division of 10 by 5"
subject: "Computer Organization and Architecture"
chapter: 5
key: "semester_5/computer_organization_and_architecture:nonrestoring-10-5"
source: https://www.bctengineer.com/q/semester_5/computer_organization_and_architecture/nonrestoring-10-5
site: BCT Engineering
---

# Non-restoring division of 10 by 5

Computer Organization and Architecture · Chapter 5: Computer Arithmetic

- Asked in: 2078 Bhadra, 5 marks
- Video: [Non Restoring Division Algorithm: Flowchart and Example | COA](https://www.youtube.com/watch?v=je7RSR6w37c) — 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/nonrestoring-10-5
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_organization_and_architecture/chapter-5

## Question

Divide 10/5 using non-restoring division.

## Short answer

Non-restoring division shifts left and either subtracts M if A was positive or adds M if A was negative, then sets Q0 from the sign of A; correct the remainder by adding M if final A is negative. For 10 divided by 5, quotient is 2 (0010) and remainder is 0.

## Answer

Non-restoring division: in each step shift A,Q left; then, if A was ≥ 0 before the shift, do A = A − M, and if A was < 0, do A = A + M. Set Q0 = 1 if the new A ≥ 0, otherwise Q0 = 0. At the end, if A < 0, add M once to correct the remainder.

Dividend = 10 = 1010 (Q), divisor M = 5 = 00101, A = 0, n = 4. No restoring is done; the next step uses the sign of A.

```text
Step  Action                              A      Q
-     Initial                             00000  1010
1     Shift left A,Q                      00001  0100
      A>=0 before shift: A = A - M (=-4)  11100  0100 (Q0=0)
2     Shift left A,Q                      11000  1000
      A<0 before shift: A = A + M (=-3)   11101  1000 (Q0=0)
3     Shift left A,Q                      11011  0000
      A<0 before shift: A = A + M (=0)    00000  0001 (Q0=1)
4     Shift left A,Q                      00000  0010
      A>=0 before shift: A = A - M (=-5)  11011  0010 (Q0=0)
      Final A<0: A = A + M                00000  0010
```

**Answer: 10 ÷ 5: quotient = 2 (0010), remainder = 0**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In non-restoring division, what is done if A < 0?
   A. Shift left, then A = A - M
   B. Shift left, then A = A + M
   C. Restore A, then subtract
   D. Shift right, then A = A + M
   Answer: B. A negative A leads to shift and add in the next cycle; no restore step. (Key point: Non-restoring rule)

2. When is Q0 set to 1?
   A. When the new A >= 0
   B. When the new A < 0
   C. Always after addition
   D. Only in the last step
   Answer: A. Q0 = 1 if the result of the add/subtract is non-negative. (Key point: Quotient bit)

3. What is the final correction and result for 10 / 5?
   A. Always subtract M; quotient 5
   B. Always add M; remainder 5
   C. If final A < 0 add M; quotient 2, remainder 0
   D. No correction ever; quotient 0
   Answer: C. Final A = 11011 was negative so M is added giving 0; quotient 0010 = 2. (Key point: Final correction)

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