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The number $43$ in $2's$ complement representation is

  1. $01010101$
  2. $11010101$
  3. $00101011$
  4. $10101011$
asked in Digital Logic by Veteran (59.5k points)
edited by | 1.9k views
0
it is really very confusing question. but now i got it.
0
In 2's complement representation positive integers are represented in its normal binary form while negative numbers are represented in its 2's complement form

4 Answers

+34 votes
Best answer

$2's$ complement representation is not same as $2's$ complement of a number. In $2's$ complement representation positive integers are represented in its normal binary form while negative numbers are represented in its $2's$ complement form. So, (c) is correct here.

http://www.ele.uri.edu/courses/ele447/proj_pages/divid/twos.html

answered by Veteran (353k points)
edited by
+5 votes

answer - C

MSB of 2's complement number has a weight of -2n-1

answered by Loyal (9k points)
+4 votes

This is exactly the sort of question, which come under "silly mistakes " Category After exam. 

Reading the question carefully is very important , answer will be 1's complement of 43 , i.e. C) 

00101011.

answered by Active (2.2k points)
+1 vote

As we know 2's complement of a number vary from 

-( 2

(n-1) ) to 2(n-1) - 1

43 is between -64 to 63, so n= 7. Therefore minimum 7 bits is needed.

In 2's complement representation, for positive number the number can be written as 0 followed by its binary representation.

for negative number 1 followed by 2's complement of it's binary representation.

As 43 is a positive number so 0 followed by binary representation of 43

43 = 32+8+2+1 = 0101011

But options are in 8 bit representation so we can make 7bit number to 8bit by sign extension in 2's complement representation i.e if first bit is 1 the new leftmost bit will be 1 and if it is zero then new leftmost bit will be 0. 
so we can rewrite 43 as 00101011 which is option C.

answered by (309 points)


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