Let $A = 1111 1010$ and $B = 0000 1010$ be two $8-bit$ $2’s$ complement numbers. Their product in $2’s$ complement is
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.
60 is represented as 111100. Make it 8 bits by adding 2 leading zeroes 00111100. As negative numbers are represented in 2’s complement, therefore Its 2's complement will be 1100 0100.
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
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
There are sequence of 1's in left side delete all consecutive 1's except last one i.e;(delete four 1's from starting) This is shortcut of finding a decimal number from 2's compliment form
First write +60 in binary (using 8 bit because here all options are in 8 bits)
+60 in binary=00111100