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

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.

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

(i) (41.8125)10(41.8125)_{10} to binary

Integer part (repeated division by 2):

DivisionQuotientRemainder
41 ÷ 2201
20 ÷ 2100
10 ÷ 250
5 ÷ 221
2 ÷ 210
1 ÷ 201

Reading upwards: 41=10100141 = 101001.

Fraction part (repeated multiplication by 2):

MultiplicationResultInteger bit
0.8125 × 21.6251
0.625 × 21.251
0.25 × 20.50
0.5 × 21.01

Reading downwards: 0.8125=.11010.8125 = .1101.

Answer: (41.8125)10=(101001.1101)2(41.8125)_{10} = (101001.1101)_2

(ii) (1000)2(1000)_2 to BCD

(1000)2=8(1000)_2 = 8, and the digit 8 in BCD is 1000.

Answer: (1000)2=(1000)BCD(1000)_2 = (1000)_{BCD} (or 0000 1000 when written as two BCD digits)

(iii) (19)10(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(19)_{10} = (0100\ 1100)_{XS-3}

✓ Answer checked by Claude Opus 5.5 (high effort)

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