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+14 votes
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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 in Digital Logic by Veteran (52k points)
edited by | 3.2k views

3 Answers

+22 votes
Best answer
$A = 1111\quad 1010 = -6$
$B = 0000\quad 1010 = 10$
$A\times B = -60 =1100\quad 0100$

Correct Answer: $A$
answered by Veteran (59.7k points)
edited by
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.

+1
Can you please tell how -60 =1100 0100
+1

60 is represented as 111100. Its 2's complement will be 000100. As answer is negative we add a sign bit 1000100 and again to make it 8 bit, we extend the sign bit 1100 0100

+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 (9.7k 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
0 votes

In 2's complement form 

A = 11111010 = -128+64+32+16+8+2 = -6

B = 00001010 = 10

Now A* B = -60

-60 = -128+64+4 =11000100

Option : A

 

answered by Junior (857 points)
0
This seems somewhat understood able.
Answer:

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