Basic

level 01

learn to

make ips distinct in same network
get network address by applying subnet mask to ip

goal 01

A and B are on the same network, so their network address must match
by applying subnet mask /24 to B's ip, you get 104.98.23.0/24
any ip address within 104.98.23.{1 ~ 254} are within the same network, 104.98.23.0
so ip of A should be within the above range
except B's (104.98.23.12), because how are you gonna distinguish between same addresses

goal 02

same as goal 01 but CIDR is /16
so they will listen to any ip address within 211.191.{0.1 ~ 255.254}
same as goal01

level 02

goal 01

client A and B's subnet mask should match -> /27
there can be 8 possible network address with usable range of 30 for /27
- xxx.xxx.xxx.{0, 32, 64, 96, 128, 160, 192, 224}
- since B's last digits are 222 -> 192.168.63.192/27
A's address can be within 192.168.63.{193 ~ 221}

goal 02

same as goal 01 but CIDR is /30
64 possible network address with usable range of 2 for /30
- xxx.xxx.xxx.{0, 4, 8 ... 252}
- address can be same as goal 01 because these are seperate networks

level 03

goal 01 ~ 03

same as level 02 but with one more client
calculated network address and range: 104.198.197.0/25
- which means usable host ip range: 104.198.197.1 ~ 104.198.197.126
just don't make their ip overlap

level 04

goal 01 ~ 03

since all masks are empty -> 75.227.117.0/24 will be easiest
- host ip range: 75.227.117.{1 ~ 254}
just don't make them overlap
there will be more after having other interfaces...