+14 votes
2.4k views

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

1. $1100 0100$
2. $1001 1100$
3. $1010 0101$
4. $1101 0101$
asked
edited | 2.4k views

## 2 Answers

+22 votes
Best answer
$A = 1111\quad 1010 = -6$
$B = 0000\quad 1010 = 10$
$A\times B = -60 =1100\quad 0100$
answered by Veteran (55.6k points)
selected
0
how 11111010 is -6
+1

In A we can reject 1111 because single sign bit is sufficient.

And we know 2's complement is weighted code.

So 1010 = -23 *1 + 22 *0 + 21*1 + 20 * 0 = -6.

Here MSB is sign bit and 1 represents -ve.

+7 votes

Explanation: Here, we have

A = 1111 1010 =  – 610 (A is a 2’s complement number)

B = 0000 1010 =  1010 (B is a 2’s complement number)

A x B = – 6010 = 1 011 11002 = 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

answered by Loyal (9.2k points)
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
Answer:

+12 votes
1 answer
1
+20 votes
4 answers
2
+25 votes
10 answers
3
+18 votes
4 answers
4
0 votes
1 answer
5
+13 votes
1 answer
6
+13 votes
3 answers
7