## Solution

The encryption shows encrypting the same ciphertext with different pairs for `N` and `e`. Luckily, looking at the first encryption pair is sufficient to get the flag.

The following values are sufficient to decrypt the flag since the full factorization of `n` is known.

```python
n = 143786356117385195355522728814418684024129402954309769186869633376407480449846714776247533950484109173163811708549269029920405450237443197994941951104068001708682945191370596050916441792714228818475059839352105948003874426539429621408867171203559281132589926504992702401428910240117807627890055235377744541913
e = 114194
ciphertext = 0x0c55bc89e3773d8e378121eced4f9300103a8696bc3f9a1542c5b1539442ca5de03a40ad564ab5c2e764b2f946058ec220abf20afc271896ff4ca1f4a2dd227405f221de51e097d6b9f270c4561cd25596e96efd7de1a0e65d37cbf6a73c62a7e323f48450b9dc75e3e738ec1c7e1ae9fc808da8c476e72aea9155125b815653
```

__For full implementation and solution see the URL__

Original writeup (https://github.com/pberba/ctf-solutions/tree/master/20181025_bsidesdelhi/crypto-100-rsababy).