First, read the linked post.

Don't worry. I understand the value of preparing your RFC 1149 network after the shipwreck.

You don't need to convert the whole IP to binary. The first two octets you already know because the mask is all ones (255 = 11111111), so 138.248.something.

The third octet you do need to convert to binary up until the end of the ones since you are doing a bitwise AND. Convert the netmask's octet to binary: 192 = 11000000. Here you only need to examine the first two bits of the third octet since the rest you know is zero. 184 is less than 192, so it must start with 10 since if it started with 11 it would have to be 192 or greater. Thus, the third octet of the network is 128.

The fourth is 0 since the fourth octet of the netmask is 0.

138.248.128.0

`ipcalc`

can help you check your math.

```
$ ipcalc 138.248.184.17/18
Address: 138.248.184.17 10001010.11111000.10 111000.00010001
Netmask: 255.255.192.0 = 18 11111111.11111111.11 000000.00000000
Wildcard: 0.0.63.255 00000000.00000000.00 111111.11111111
=>
Network: 138.248.128.0/18 10001010.11111000.10 000000.00000000
HostMin: 138.248.128.1 10001010.11111000.10 000000.00000001
HostMax: 138.248.191.254 10001010.11111000.10 111111.11111110
Broadcast: 138.248.191.255 10001010.11111000.10 111111.11111111
Hosts/Net: 16382 Class B
```

asan answer, but I'd make it just a simple statement: "Read this post." – Ward Mar 2 '11 at 22:57