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\section*{Problem 3}
+\paragraph{First solution \emph{(too easy, feels like cheating).}} We saw in
+class how to compute the most signigicant bit set to one. But observe that
+\texttt{x \& $\sim$(x-1)} will clear all the bits of $x$ except for the least
+significant bit set to one. Assuming access to a procedure \texttt{msb} to
+compute the most significant bit set to one, we can simply define
+\texttt{lsb(x) $:=$ msb(x \& $\sim$(x-1))}.
+
+\paragraph{Second solution.}
Using Problem 2 and what we did in class, we assume that we have access to the
following procedures:
\begin{itemize*}