Explanation:
\(11(p+110)=1001 p\)
we multiply in base two
\(11 p+10010=1001 p\)
\(10010=1001 p-11 p\)
\(10010=110 p(\) subtract ion in base 2 ).
Then, we convert to base 10
\begin{aligned}
&1 \times 2+1 \times 21=(1 \times 22+1 \times 2) p \\
&16+2=6 p \\
&\Rightarrow p=\frac{18}{6}=3
\end{aligned}
\begin{array}{l|lll}
2 & 3 & & \\
\hline 2 & 1 & r & 1 \\
0 & r & 1
\end{array} \mid
Then. \(P\) in base two equals \(11_{2}\)