Network

Challenge

Combien d'ip maximum contient un réseau /18

Quelle est l'IP de broadcast du réseau 172.16.0.0/23 (la réponse est la concaténation de la partie hôte. Exemple pour 192.168.0.0/16 ---> 255255)

Si je possède 8 sous réseau dans le réseau 192.168.0.0/24. Quelle est la première IP du 5e réseau ? (Uniquement le dernier block)

1111010

Le flag est la concaténation des 4 réponses séparé par un point ctf{1.2.3.4}

Solution

This task contained 4 different things to calculate.

First, a /18 network is calculated by substracting the mask with 32 (which is the maximum amount of bits an IP address can have). So 32-18=14 and 2 power 14 = 16384.

Second, an IP address in /23 means 512 hosts and a subnet mask of 255.255.254.0. So the range goes from 172.16.0.1 to 172.16.1.254 and that means the broadcast IP is 172.16.1.255.

Third, if the network is 192.168.0.0/24 that means the range is 192.168.0.1 to 192.168.0.254 and the total number of hosts is 256. Considering we have to split it into 8 subnets that means every network contains 32 IP addresses. So we count:

Subnet	Start	End
1	192.168.0.0	192.168.0.31
2	192.168.0.32	192.168.0.63
3	192.168.0.64	192.168.0.95
4	192.168.0.96	192.168.0.127
5	192.168.0.128	192.168.0.159

Fourth, convert the binary numbers into decimal which gives us 122.