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If $x=(1101)_{2}$ and $y=(110)_{2}$, then what is the value of $x^{2}-y^{2}$ ?
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The correct answer is:
$\left(\begin{array}{llllll}1 & 0 & 0 & 0 & 0 & 1 & 0 & 1\end{array}\right)_{2}$
Given, $x=(1101)_{2}=1 \times 2^{3}+1 \times 2^{2}+0 \times 2^{1}+1$
$=8+4+1=13$
and $\begin{aligned} y=(110)_{2}=& 1 \times 2^{2}+1 \times 2^{1}+0 \times 2^{0} \\=4+2=6 \end{aligned}$
$\therefore x^{2}-y^{2}=(13)^{2}-(6)^{2}=169-36=133$
Now, \begin{array}{l|l|l}
2 & 133 & \\
\hline 2 & 66 & 1 \\
\hline 2 & 33 & 0 \\
\hline 2 & 16 & 1 \\
\hline 2 & 8 & 0 \\
\hline 2 & 4 & 0 \\
\hline 2 & 2 & 0 \\
\hline & 1 & 0
\end{array}
$\therefore 133=(10000101)_{2}$
$=8+4+1=13$
and $\begin{aligned} y=(110)_{2}=& 1 \times 2^{2}+1 \times 2^{1}+0 \times 2^{0} \\=4+2=6 \end{aligned}$
$\therefore x^{2}-y^{2}=(13)^{2}-(6)^{2}=169-36=133$
Now, \begin{array}{l|l|l}
2 & 133 & \\
\hline 2 & 66 & 1 \\
\hline 2 & 33 & 0 \\
\hline 2 & 16 & 1 \\
\hline 2 & 8 & 0 \\
\hline 2 & 4 & 0 \\
\hline 2 & 2 & 0 \\
\hline & 1 & 0
\end{array}
$\therefore 133=(10000101)_{2}$
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