2076 Chaitra · 1.5 marks

Convert (17)excess-3 to BCD

Short answer

Assuming decimal 17, its excess-3 is 0100 1010. Subtract 0011 from each nibble to convert to BCD: 0100−0011=0001, 1010−0011=0111 → 0001 0111 (BCD).

Convert (17)excess-3 = (?)BCD.

"17" is not a valid excess-3 bit pattern (XS-3 is written in 4-bit binary groups), so the question is read as: write decimal 17 in excess-3 code and in BCD.

BCD: each decimal digit in 4-bit binary.

Digit	1	7
BCD	0001	0111

Excess-3: each decimal digit + 3, in 4-bit binary.

Digit	1	7
Digit + 3	4	10
XS-3	0100	1010

Going from XS-3 back to BCD means subtracting 0011 from each group: $0100 - 0011 = 0001$ , $1010 - 0011 = 0111$ .

Answer: $(17)_{10} = (0100\ 1010)_{XS\text{-}3} = (0001\ 0111)_{BCD}$

✓ Answer checked by Claude Opus 5.5 (high effort)

Convert (1100 0011)Excess-3 to BCD · Digital Logic, chapter 1
Convert (1001 1010)excess-3 to BCD · Digital Logic, chapter 1
Convert (1010 0111)excess-3 to BCD · Digital Logic, chapter 1
Gray→BCD, decimal→Excess-3 and 2's complement addition · Digital Logic, chapter 1
(1110)gray→BCD and (1430)₁₀→Excess-3 · Digital Logic, chapter 1
Digital system advantages; BCD and Excess-3 · Digital Logic, chapter 1
8085 ALP: BCD to binary conversion · Microprocessor, chapter 2

All Digital Logic chapter 1 questions →

Convert (17)excess-3 to BCD

Related questions