Minimum number of bits to represent negative number
The representations of the numbers as 6-bit two's complement binaries are
\begin{align}
+32_{10} &amp;= 10\,0000_2\\
-32_{10} &amp;= \overline{10\,0000}_2 + 1_2\\
         &amp;= 01\,1111_2 + 1_2\\
         &amp;= 10\,0000_2
\end{align}
which would give the same representation for both numbers. That means you cannot use 6-bits numbers to represent both values using two's complement.

For 7-bit binaries it works, however:
\begin{align}
+32_{10} &amp;= 010\,0000_2\\
-32_{10} &amp;= \overline{010\,0000}_2 + 1_2\\
         &amp;= 101\,1111_2 + 1_2\\
         &amp;= 110\,0000_2
\end{align}
As you can see, the most significant bit represents the sign of the number.



Another perspective on your question: What is the biggest positive/smallest negative number you can store in an $n$-bit two's complement binary? Then the answer is $2^{(n-1)}-1$ for the biggest positive and $-2^{(n-1)}$ for the smallest negative number. 

That means that a 6-bit binary can hold values $-32, \ldots, 31$ and a 7-bit binary values $-64, \ldots, 63$. The number $-32$ then fits into the range of 6-bit binaries, while $+32$ doesn't. You'd need one more bit for the latter.