Subnetting in Your Head — The Magic-Number Trick (CCNA)

If subnetting feels like binary-math homework every time, you’re doing it the slow way. There’s a trick — the magic number — that turns the whole thing into a 10-second mental calculation. Once you have it, you’ll never go back to writing out 32-bit binary on scratch paper.

The idea

Every subnet mask “splits” one octet of the IP address. The size of each subnet’s block, in that octet, is the magic number: 256 − the mask value of that octet.

That’s the whole trick.

Walk through /27

A /27 mask is 255.255.255.224. The interesting octet is the fourth (the only one that isn’t 0 or 255).

Magic number = 256 − 224 = 32.

So subnets in a /27 are 32 addresses wide. Networks land at multiples of 32:

192.168.10.0  /27   →  hosts .1   – .30,   broadcast .31
192.168.10.32 /27   →  hosts .33  – .62,   broadcast .63
192.168.10.64 /27   →  hosts .65  – .94,   broadcast .95
192.168.10.96 /27   →  hosts .97  – .126,  broadcast .127

Given an IP like 192.168.10.50/27 — what’s its subnet?

Magic number is 32. The largest multiple of 32 that’s ≤ 50 is 32. So the subnet is 192.168.10.32/27. Broadcast is 192.168.10.63. Usable hosts are .33 through .62. Done in 5 seconds, no binary.

Walk through /28

Mask 255.255.255.240. Magic number = 256 − 240 = 16.

Subnets are 16 addresses wide: .0, .16, .32, .48, .64, … .240.

192.168.10.45/28 → biggest multiple of 16 ≤ 45 is 32. Subnet = 192.168.10.32/28. Broadcast = 192.168.10.47. Hosts .33 – .46.

Walk through /22

Now the interesting octet is the third (because /22 borrows into octet 3).

Mask 255.255.252.0. Magic number = 256 − 252 = 4.

Subnets are 4 addresses wide in the third octet: 0.0, 4.0, 8.0, 12.0, …

172.16.10.0/22 → biggest multiple of 4 ≤ 10 is 8. Subnet = 172.16.8.0/22. The full range covers 172.16.8.0 through 172.16.11.255. Broadcast = 172.16.11.255. That’s 1,024 addresses, 1,022 usable hosts.

Walk through /21

Mask 255.255.248.0. Magic number = 256 − 248 = 8.

Subnets are 8 addresses wide in the third octet: 0.0, 8.0, 16.0, 24.0, …

10.50.20.5/21 → biggest multiple of 8 ≤ 20 is 16. Subnet = 10.50.16.0/21, range 10.50.16.0 – 10.50.23.255. Broadcast = 10.50.23.255.

The whole process — 5 steps

1. Identify the interesting octet (the one where the mask isn’t 0 or 255).
2. Compute the magic number: 256 − mask value of that octet.
3. Find the biggest multiple of the magic number that’s ≤ the IP’s value in that octet → that’s your subnet’s network address for that octet.
4. Broadcast = network address + magic number − 1.
5. Usable hosts = network address + 1 through broadcast − 1.

That’s it. The same five steps work for every mask between /9 and /30.

Memorize the mask table

Don’t compute masks every time — memorize the 8 values:

Bits	Mask value
1	128
2	192
3	224
4	240
5	248
6	252
7	254
8	255

A /27 mask in the fourth octet has 3 borrowed bits → look up “3” → 224. /22 mask in the third octet has 6 borrowed bits → “6” → 252.

How many subnets / how many hosts?

• Subnets = 2^(borrowed bits). /27 from a /24 = 3 borrowed bits = 8 subnets.
• Hosts per subnet = 2^(host bits) − 2. /27 has 5 host bits = 2^5 − 2 = 30 hosts.

These are the only formulas you need. Combined with the magic-number trick, you can answer any CCNA subnetting question in under a minute.

A quick drill

Find subnet, broadcast, and usable host range for each:

1. 192.168.20.85/29
2. 10.1.1.100/26
3. 172.20.5.5/23
4. 192.168.0.200/27

Try it before looking. Answers below.

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Answers:

1. /29 → magic 8. 85 → biggest mult of 8 ≤ 85 = 80. Subnet 192.168.20.80/29, broadcast .87, hosts .81–.86.
2. /26 → magic 64. 100 → biggest mult of 64 ≤ 100 = 64. Subnet 10.1.1.64/26, broadcast .127, hosts .65–.126.
3. /23 → mask 255.255.254.0. Third octet interesting, magic = 256 − 254 = 2. 5 → biggest mult of 2 ≤ 5 = 4. Subnet 172.20.4.0/23, broadcast 172.20.5.255, hosts 172.20.4.1 – 172.20.5.254.
4. /27 → magic 32. 200 → biggest mult of 32 ≤ 200 = 192. Subnet 192.168.0.192/27, broadcast .223, hosts .193–.222.

If you got 3 of 4 right, you’ve got the technique. If you missed any, you probably forgot to identify the interesting octet first — that’s the #1 mistake.

Where to go from here

If you want more — the full Subnetting library topic covers the binary-math version (slower but more rigorous), VLSM (variable-length masks for efficient address allocation), and the “I have a /24 — give me four equal subnets” pattern. The magic-number trick covers 95% of what you’ll see on the CCNA exam.