n = 1763350599372172240188600248087473321738860115540927328389207609428163138985769311

e = 65537

c = 33475248111421194902497742876885935310304862428980875522333303840565113662943528

- looks like the problem is asking to exploit a weak key to find the flag.
- did some googling and found [a tool for this very purpose](https://github.com/Ganapati/RsaCtfTool)

`python3 RsaCtfTool/RsaCtfTool.py -n 1763350599372172240188600248087473321738860115540927328389207609428163138985769311 -e 65537 --uncipher 33475248111421194902497742876885935310304862428980875522333303840565113662943528`

Output:

Unciphered data :
HEX : 0x00000000007368656c6c7b737769746368696e5f746f5f6173796d6d65747269637d
INT (big endian) : 3111388068276188662361997958100924356274395167698926770307665056326525
INT (little endian) : 3716857967501616239523840250653395077772235796196542527851123201402003116282347520
STR : b'\x00\x00\x00\x00\x00shell{switchin_to_asymmetric}'

**flag: shell{switchin_to_asymmetric}**

Original writeup (https://github.com/ivanchubb/CTF-Writeups/blob/main/2021/S.H.E.L.L.%20CTF/Easy%20RSA.md).