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Number conversions (903)10 to BCD etc.

Short answer

a) 903 BCD: 1001 0000 0011. b) Octal 624.03 to hex via binary: 194.0C. c) 101101+10111 = 1000100 (68). d) 2’s complement of 11 is 0101; 14+0101=0011 (3).

Perform the following: a) (903)10 = (?)BCD b) (624.03)8 = (?)16 c) (101101)2 + (10111)2 = ? d) Subtract using 2's complement: (14)10 − (11)10

a) (903)₁₀ to BCD

9 = 1001, 0 = 0000, 3 = 0011

(903)₁₀ = (1001 0000 0011)BCD

b) (624.03)₈ to hexadecimal

624 = 110 010 100
.03 = 000 011
Integer: 1 1001 0100  -> 1 9 4
Fraction: 0000 1100   -> 0 C

(624.03)₈ = (194.0C)₁₆

c) (101101)₂ + (10111)₂

   101101    (45)
 + 010111    (23)
 --------
  1000100    (68)

Carries are propagated from the LSB: 1+1 = 10, etc.

Answer: 1000100₂ (= 68₁₀)

d) (14)₁₀ − (11)₁₀ using 2's complement (4 bits)

  • 14 = 1110; 11 = 1011
  • 2's complement of 1011 = 0100 + 1 = 0101
   1110
 + 0101
 ------
 1 0011   discard the end carry

End carry present, so the result is positive: 0011.

Answer: 14 − 11 = (0011)₂ = 3

✓ Answer checked by Claude Opus 5.5 (high effort)

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