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When two $4-bit$ numbers $A = a_3a_2a_1a_0$ and $B=b_3b_2b_1b_0$ are multiplied, the bit $c_1$ of the product $C$ is given by ________
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+41 votes
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$\begin{matrix}
& & & &a_3 & a_2 & a_1&a_0 \\
 & & &\times & b_3 &b_2 &b_1 &b_0 \\
\hline
& & & & a_3b_0 &a_2b_0 &a_1b_0 & a_0b_0&\\
& & &a_3b_1 &a_2b_1 &a_1b_1 & a_0b_1 &- &\\
& &a_3b_2 &a_2b_2 &a_1b_2 &a_0b_2 &- &- &\\
& a_3b_3 &a_2b_3 &a_1b_3 &a_0b_3 &- &- &- \\
\hline
c_7&c_6&c_5 &c_4 &c_3 &c_2 &c_1 &c_0 \\ \hline
\end{matrix}$

$c_1=b_1 a_0 \oplus a_1 b_0$
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+3
how?
0
corrected...
0
Yes. It must be symmetric :)
+1
it must be a1b0 and b1a0.
0
PLEASE GIVE SOLUTION?
0
i m not able to understand why xor is applied here
+41
When we consider 2 bits, ADD and XOR works the same except when both inputs are 1. In this case, ADD gives 10, while XOR gives 0. Since, we are asked only about the bit c1, our answer should be 0 which is given by XOR.
+1
What would be the answer if instead of c1  they are asking for  c2  ?  Plz explain ?
0
why xor , not getting :(
0
@ashwina

I think ans would be same.
+8

sid1221  We need to add these two bits,so we can use same expression of sum of half adder which is exor

0
oh yes got it thanks :)
+5
I don't think expression for C1 and C2 will be same. For C2 we will need a full adder.
0
yes for C2 we would be needing a full adder
0

@VS

I think expression for C2 will not be same ..value of C2 depend on carry if generated while calculating C1,

Hence C2 => C1 $\bigoplus$ Carry out of First adder

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