- List each address in binary format and determine the common prefix for all of the addresses:
$$
\begin{aligned}
& 212.56 .146 .0 / 24:\;\;11010100.00111000 .10010010 .00000000 \\
& 212.56 .147 .0 / 24:\;\;11010100.00111000 .10010011 .00000000 \\
& 212.56 .148 .0 / 24:\;\;11010100.00111000 .10010100 .00000000 \\
& 212.56 .149 .0 / 24:\;\;11010100.00111000 .10010101 .00000000
\end{aligned}
$$
- Note that this set of four $/ 24 \mathrm{~s}$ cannot be summarized as a single $/ 23$.
$$
\begin{aligned}
& 212.56 .146 .0 / 23:\;\;11010100.00111000 .10010010 .00000000 \\
& 212.56 .148 .0 / 23:\;\;11010100.00111000 .10010100 .00000000
\end{aligned}
$$
- The CIDR aggregation is:
$$
\begin{aligned}
& 212.56 .146 .0 / 23 \\
& 212.56 .148 .0 / 23
\end{aligned}
$$
Note that if two $/ 23 \mathrm{~s}$ are to be aggregated into a $/ 22$, then both $/ 23 \mathrm{~s}$ must fall within a single $/ 22$ block. Since each of the two $/ 23 \mathrm{~s}$ is a member of a different $/ 22$ block, they cannot be aggregated into a single $/ 22$ (even though they are consecutive). They could be aggregated into $222.56.144/21,$ but this aggregation would include four network numbers that were not part of the original allocation. Hence, the smallest possible aggregate is two $/ 23 \mathrm{~s}$.
