Tags: modular-arithmetic
Rating: 0
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,
```python
inve = pow(e, -1, n)
dec = (inve * enc) % n
flag = int.to_bytes(dec, byteorder='big')
```
for more info check the original writeup
https://github.com/TalaatHarb/ctf-writeups/tree/main/irisctf2023/babynotrsa
