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

1. $01010101$
2. $11010101$
3. $00101011$
4. $10101011$

it is really very confusing question. but now i got it.
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

Positive numbers look same in Sign-Magnitude, 1's Complement, 2's Complement representation.

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$2's$ 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 or archive

by

nice point arjun sir

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 true magnitude of 43 , i.e. C)

00101011.

### 1 comment

Key point

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

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.