Rating:
# Find Me
The given file was a PCAP capture. The first thing we do is `strings` to have easy informations.
![strings](https://github.com/Ewael/CTFs/blob/master/2021/ShaktiCTF/Forensics/FindMe/strings.png?raw=true)
We see this interesting part:
```
You might need this...
98 106 66 48 100 71 103 48 100 71 86 104 78 88 107 61
```
Let's decode this:
```python
>>> print(''.join([chr(int(a)) for a in '98 106 66 48 100 71 103 48 100 71 86 104 78 88 107 61'.split(' ')]))
bjB0dGg0dGVhNXk=
```
```
$ echo bjB0dGg0dGVhNXk= | base64 -d
n0tth4tea5y
```
Ok, let's keep this and dive into the capture file. Following the TCP streams gives us 2 streams, the first one being what we just decoded, let's dig the second one.
![cap1](https://github.com/Ewael/CTFs/blob/master/2021/ShaktiCTF/Forensics/FindMe/cap1.png?raw=true)
![cap2](https://github.com/Ewael/CTFs/blob/master/2021/ShaktiCTF/Forensics/FindMe/cap2.png?raw=true)
We can clearly see a reversed `flag.txt` string in that capture, let's reverse it and see we get as a file.
![cyber](https://github.com/Ewael/CTFs/blob/master/2021/ShaktiCTF/Forensics/FindMe/cyber.png?raw=true)
A ZIP file! It's password protected, opening it with our `n0tth4tea5y` works like a charm, and we get out beloved `flag.txt`:
```
shaktictf{g00d_lUcK_4_tH3_n3xT_cH411eNg3}
```