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The 15's complement of A9BH is :

  • A.\(\mathrm{B02_{H}}\)
  • B.\(\mathrm{173_{H}}\)
  • C.\(\mathrm{CD2_{H}}\)
  • D.\(\mathrm{564_{H}}\)

Answer: D

The 15's complement of A9BH is 564H.

The 15's complement in hexadecimal is the analogue of the 1's complement in binary: subtract every digit from F (15).

\[ \begin{aligned} F - A &= 15 - 10 = 5 \\ F - 9 &= 15 - 9 = 6 \\ F - B &= 15 - 11 = 4 \end{aligned} \]

So the result is 564H. As a check, A9B + 564 = FFF. Adding 1 to the 15's complement would give the 16's complement, 565H, which is not offered; the other options do not add to FFF with A9B.

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