---
title: "Code conversions: (41.8125)₁₀, (1000)₂, (19)₁₀"
subject: "Digital Logic"
chapter: 1
key: "semester_3/digital_logic:code-conversions-41-8125"
source: https://www.bctengineer.com/q/semester_3/digital_logic/code-conversions-41-8125
site: BCT Engineering
---

# Code conversions: (41.8125)₁₀, (1000)₂, (19)₁₀

Digital Logic · Chapter 1: Introduction

- Asked in: 2078 Kartik, 2+2+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/digital_logic/code-conversions-41-8125
- Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-1

## Question

Perform the following code conversions: (i) (41.8125)₁₀ = (?)₂ (ii) (1000)₂ = (?)BCD (iii) (19)₁₀ = (?)Ex-3 code

## Short answer

(i) 41.8125₁₀ to binary: integer 41 = 101001 by division, fraction 0.8125 = .1101 by repeated multiplication, giving 101001.1101. (ii) (1000)₂ = 8₁₀ → BCD 1000. (iii) (19)₁₀ to Excess‑3: 1+3=4→0100, 9+3=12→1100 → 0100 1100.

## Answer

#### (i) $(41.8125)_{10}$ to binary
**Integer part** (repeated division by 2):

| Division | Quotient | Remainder |
|---|---|---|
| 41 ÷ 2 | 20 | 1 |
| 20 ÷ 2 | 10 | 0 |
| 10 ÷ 2 | 5 | 0 |
| 5 ÷ 2 | 2 | 1 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 |

Reading upwards: $41 = 101001$.

**Fraction part** (repeated multiplication by 2):

| Multiplication | Result | Integer bit |
|---|---|---|
| 0.8125 × 2 | 1.625 | 1 |
| 0.625 × 2 | 1.25 | 1 |
| 0.25 × 2 | 0.5 | 0 |
| 0.5 × 2 | 1.0 | 1 |

Reading downwards: $0.8125 = .1101$.

**Answer: $(41.8125)_{10} = (101001.1101)_2$**

#### (ii) $(1000)_2$ to BCD
$(1000)_2 = 8$, and the digit 8 in BCD is 1000.

**Answer: $(1000)_2 = (1000)_{BCD}$** (or 0000 1000 when written as two BCD digits)

#### (iii) $(19)_{10}$ to Excess-3
Add 3 to each digit: 1 + 3 = 4 → 0100; 9 + 3 = 12 → 1100.

**Answer: $(19)_{10} = (0100\ 1100)_{XS-3}$**

## Check yourself

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

1. What is 41 decimal in binary?
   A. 100101
   B. 101010
   C. 101001
   D. 110001
   Answer: C. 41 = 32 + 8 + 1 = 101001. (Key point: Integer method)

2. How is the fraction 0.8125 converted to binary?
   A. Repeated division by 2, reading upward
   B. Multiplying by 10
   C. Dividing by 16
   D. Repeated multiplication by 2, reading integer parts downward
   Answer: D. 0.8125 gives bits 1,1,0,1 = .1101. (Key point: Fraction method)

3. What is (1000) binary in BCD?
   A. 1000 (the digit 8)
   B. 0001 0000
   C. 1000 0000
   D. 0111
   Answer: A. 1000 binary is 8, and BCD of 8 is 1000. (Key point: Binary to BCD)

4. What is (19) decimal in Excess-3?
   A. 0001 1001
   B. 0100 1100
   C. 0100 1001
   D. 0011 1100
   Answer: B. 1+3=4 gives 0100; 9+3=12 gives 1100. (Key point: XS-3 of 19)

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