Networking
Networking challenge section
IPV4
In this section we are given pcap file which is Wireshark file. Let's analysis this file.
In this file we found at packet comment section, there is base64 encoded text. Could be a flag, let's try to decode it.
I got the flag after decode the base64.
EG{W1R3SH4RK_1S_GR34T}IPV6
In this challenge, we're given a pcap file again to analysis it. Let's do it.
At first, I did not find anything in the pcap file. Stuck for a while. Then I remember we can scan or analysis the pcap file using online tools on internet to find hidden message. I use this online tools to scan the pcap file https://filext.com/online-file-viewer.html.
Decode the base64 encode: RUd7VEgxU18xU19USDNfQzBSUjNDVF9GTDRHfQ== and you will get the flag.
EG{TH1S_1S_TH3_C0RR3CT_FL4G}Airdream Network
In this challenge, we will learn how to calculate a host and what class is the network is.
We need to download flag.zip to get the file, but unfortunately the file is required password.
Don't forget to click reward to know password format for the zip file.
For this challenge, this network class C because the IP is start from 192.
Let's calculate number of host, subnet and hosts available. First for IP 192.168.1.0 has a default subnet mask is 255.255.255.240. So, because of our IP class is C class, so there will be 3 first octets which is 11111111.11111111.11111111.11110000. The rest of subnet bit will be 11110000. So this is our subnet bit, which is 4. So our subnets = 4.
Then let's calculate amount of subnet, the formula is 2^n. Which is n= subnet, which is 4. So 2^4 =16. This is the amount of our subnet, which is 16.
Then final step is calculated the number of host. The formula is 2^4-2. Why -2? It is because it already reserves by network address and broadcast address. So after -2 the answers become 14. Our host is 14.
After we arrange the password, it will become like this. And you will get the flag after open the zip file.
EG{1P_SUBN3TT1NG_1S_34SY}Last updated