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+14 votes

Let $A = 1111 1010$ and $B = 0000 1010$ be two $8-bit$ $2’s$ complement numbers. Their product in $2’s$ complement is

- $1100 0100$
- $1001 1100$
- $1010 0101$
- $1101 0101$

+22 votes

Best answer

+7 votes

**Explanation:** Here, we have

A = 1111 1010 = – 6_{10} (A is a 2’s complement number)

B = 0000 1010 = 10_{10} (B is a 2’s complement number)

A x B = – 60_{10} = 1 011 1100_{2} = 1 100 0011 (1’s complement) = 1 100 0100 (2’s complement)

Thus, the product of A and B in 2’s complement is 1100 0100, which is option A.

So, A is the correct option

0

for this question wht I did first,simply multiply both numbers and the result which I got 000100111000100

after taking 2's complement of this -2500 came.plz someone verify this.

I long it is taking long time but I want to know my mistake [email protected] @debashish

after taking 2's complement of this -2500 came.plz someone verify this.

I long it is taking long time but I want to know my mistake [email protected] @debashish

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