Exercise 2: Splitting 172.16.0.0/24 Into 8 Equal Subnets — Possible Solution
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Determining the new prefix:

Per the chapter's own worked example, splitting a network into N equal
subnets means borrowing enough additional bits from the host portion
so that 2^(borrowed bits) = N. To get 8 subnets, borrowed bits must
satisfy 2^x = 8, so x = 3 bits need to be borrowed.

Starting from /24 and borrowing 3 more bits: 24 + 3 = /27.

This matches the chapter's own compare-table entry for /27: 32 total
addresses per subnet, 30 usable hosts each -- consistent with taking a
256-address /24 block and splitting it into 8 equal pieces
(256 / 8 = 32 addresses per piece, matching /27's own 32-address block
size exactly).

The 8 resulting subnets:

Each subnet is 32 addresses wide (per /27's own block size), so the
third octet stays at 0 (still within the original /24) and the fourth
octet increases by 32 for each successive subnet:

172.16.0.0/27      (addresses .0   - .31)
172.16.0.32/27     (addresses .32  - .63)
172.16.0.64/27     (addresses .64  - .95)
172.16.0.96/27     (addresses .96  - .127)
172.16.0.128/27    (addresses .128 - .159)
172.16.0.160/27    (addresses .160 - .191)
172.16.0.192/27    (addresses .192 - .223)
172.16.0.224/27    (addresses .224 - .255)

Each of these 8 blocks covers exactly 32 addresses (30 usable hosts,
after subtracting each subnet's own network and broadcast address),
and together they account for the entire original 256-address
172.16.0.0/24 range with no gaps and no overlap.

WHY THIS WORKS AS AN ANSWER
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This derives the new prefix mathematically (2^x = 8 subnets needed →
3 bits borrowed → /27) rather than guessing, and lists all 8 subnets
using the correct block size (32) with no gaps or overlaps, verifying
the total against the original /24's own 256 addresses.