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+12 votes
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The smallest integer that can be represented by an $8-bit$ number in $2's$ complement form is

  1. $-256$
  2. $-128$
  3.  $-127$
  4. $0$
asked in Digital Logic by Veteran (352k points)
edited by | 2.4k views

2 Answers

+23 votes
Best answer

Range of 2's compliment no = > (- 2n-1 )to + (2n-1 - 1 )

Here n = No of bits = 8.

So minimum no = -2 ^ 7 = (B) -128

answered by Boss (42.5k points)
selected by
+2 votes
answer - B
answered by Loyal (9k points)


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