Using the **multiplicative modular inverse** of e and multiplying it with the encrypted flag should result in a decrypted flag, that can be converted into a string to get the flag,
```pythoninve = pow(e, -1, n)dec = (inve * enc) % nflag = int.to_bytes(dec, byteorder='big')```
for more info check the original writeuphttps://github.com/TalaatHarb/ctf-writeups/tree/main/irisctf2023/babynotrsa