# (un)F0r7un@t3
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```txtThree of our best agents: Ron, Adi, and Leonard, were investigating the notorious hacker group known as the unF0r7un@t3s. Fortunately, one of their members dropped a fortune cookie behind from their last scandal. We believe it might be a clue to where they will attack next. The fortune inside read: You will find your fortune at 23441987, 31018357. The lucky numbers on the back were, n52035749, e101, and *10^-6. What location do you think they will be heading to next?
NOTE: No wrapper is necessary for this problem.
- YoshiBoi#2008```
---
Ok... several hints in the task description ^^. First of all, the names of the best agents `Ron`, `Adi` and `Leonard`, coincidentally are the firstnames of the authors of the _RSA_ crypto system. Also the names of the parameters: `n` and `e` also exist in _RSA_.
So... considering this, we concluded that the first step of getting the flag was cracking the encrypted RSA.
You can simply use [RSACtfTool](https://github.com/Ganapati/RsaCtfTool) for this:
```bashpython27 RsaCtfTool.py -e101 -n52035749 --uncipher 23441987```
```txtUnciphered data :HEX : 0x02e98cd7INT (big endian) : 48860375INT (little endian) : 3616336130STR : b'\x02\xe9\x8c\xd7'```
```bashpython27 RsaCtfTool.py -e101 -n52035749 --uncipher 31018357```
```txtUnciphered data :HEX : 0x23aefeINT (big endian) : 2338558INT (little endian) : 16690723STR : b'#\xae\xfe'HEX : 0x0023aefeINT (big endian) : 2338558INT (little endian) : 4272825088STR : b'\x00#\xae\xfe'```
... all that was left to do now, was to take the `INT (big endian)` results, divide them by the last lucky number, and enter the resulting GPS coordinates on Google Maps:
![Louvre](./louvre.png)
... the flag therefore was: `Louvre`