• 192.168.204.0/24
• 192.168.205.0/24
• 192.168.206.0/24
• 192.168.207.0/24
• 192.168.208.0/24

First I will write out the binary form of the addresses up to and including the changing octet. 11000000.10101000.11001100

11000000.10101000.11001101

11000000.10101000.11001110

11000000.10101000.11001111

11000000.10101000.11010000

From the list, I counted from the left how many bits are the same in each address, as can be seen the first 19 bits for each address are the same so that gives us the subnet mask in slash notation. So the summarised address and subnet mask is 192.168.204.0/19.

Would this be correct?

-

One incorrect step:

11000000.10101000.110 is the common part. You then revert it to binary notation and get 192.168.192.0/19.

Basically, X.X.X.0/19 means that there are 2^(32-19) = 2^13 IP addresses in given subnetwork. As a rule, subnetworks cannot intersect. So for 19 bit mask, you have:

• 192.168.0.0/19
• 192.168.32.0/19
• 192.168.64.0/19
• 192.168.96.0/19
• 192.168.128.0/19
• 192.168.160.0/19
• 192.168.192.0/19
• 192.168.224.0/19

Total 8 /19 networks in 192.168.0.0/16 family (19-16=3, 2^3=8).

-
thank you, done a few practise ones online there using your format and got them all correct thank you. –  user68062 Jun 1 '11 at 20:16