The key is to look at the start and end addresses in binary, specifically focusing on the octet where the range begins and ends (the third octet in this case).
- Start Address (third octet): $8=00001000$
- End Address (third octet): $15=00001111$
When you compare them, you can see the first 5 bits ( 00001 ) are identical. The rest of the bits differ, which defines the range for the host addresses.
The prefix length is the number of constant bits:
- 8 bits (from 172 ) +8 bits (from 16 ) +5 bits (from the third octet) $=21$ bits.
- This gives us a prefix of $\mathbf{/ 2 1}$.
The network address is always the first address in the range. In this case, the range is defined to start at $172.16 .8 .0$ , which perfectly matches the prefix we found.
Combining the network address and the prefix length gives you the final answer:
$$
172.16 .8 .0 / 21 \text {. }
$$
