In this challenge we got 3 files
1- python code
```python# no secrets for you!flag = ...
# Prime numbersp = 151974537061323957822386073908385085419559026351164685426097479266890291010147521691623222013307654711435195917538910433499461592808140930995554881397135856676650008657702221890681556382541341154333619026995004346614954741516470916984007797447848200982844325683748644670322174197570545222141895743221967042369q = 174984645401233071825665708002522121612485226530706132712010887487642973021704769474826989160974464933559818767568944237124745165979610355867977190192654030573049063822083356316183080709550520634370714336131664619311165756257899116089875225537979520325826655873483634761961805768588413832262117172840398661229n = p * q
# a public exponent hidden away by Windy's musical talentse = ...
# Converting the message to an integerm = int.from_bytes(message.encode(), 'big')
# Encrypting the message: c = m^e mod ninc_m = pow(message_int, e, n)
print(encrypted_message_int)```
2-this image
![](https://rbih-boulanouar.github.io/UofTCTF-2024/Cryptography/Pianoman/musical_e.png)
3-the output
```13798492512038760070176175279601263544116956273815547670915057561532348462120753731852024424193899030774938204962799194756105401464136384387458651343975594539877218889319074841918281784494580079814736461158750759327630935335333130007375268812456855987866715978531148043248418247223808114476698088473278808360178546541128684643502788861786419871174570376835894025839847919827231356213726961581598139013383568524808876923469958771740011288404737208217659897319372970291073214528581692244433371304465252501970552162445326313782129351056851978201181794212716520630569898498364053054452320641433167009005762663177324539460```
in the python code we can see its RSA encryption we have all we need except “e” and we can see the comment say “#a public exponent hidden away by Windy’s musical talents” it seems like we can get “e” from the image.
ok, now lets see the image , its music notes it can be decode using [dcode](https://www.dcode.fr/music-sheet-cipher) e = 7029307
after decoding it we found e and we can decrypt the text using [RsaCtfTool](https://github.com/RsaCtfTool/RsaCtfTool)
uoftctf{AT1d2jMCVs03xxalViU9zTyiiV1INNJY}