An [attachment](https://static.n00bzunit3d.xyz/Crypto/RSA/encryption.txt) got us everything we need to decrypt ciphertext to plaintext again, here are what we got

e is public exponent (used for encryption)
n is modulus
c is ciphertext
Because e is 3 (small number) then if plaintext is also small enough, we can decrypt ciphertext easily by computing the integer cube root of c

and here is the python script that will do just that

```
import gmpy2

# Given values
e = 3
n = 135112325288715136727832177735512070625083219670480717841817583343851445454356579794543601926517886432778754079508684454122465776544049537510760149616899986522216930847357907483054348419798542025184280105958211364798924985051999921354369017984140216806642244876998054533895072842602131552047667500910960834243
c = 13037717184940851534440408074902031173938827302834506159512256813794613267487160058287930781080450199371859916605839773796744179698270340378901298046506802163106509143441799583051647999737073025726173300915916758770511497524353491642840238968166849681827669150543335788616727518429916536945395813

# Compute cube root
m, exact = gmpy2.iroot(c, e)

if exact:
print("The message is:", m)
else:
print("No direct cube root found, message might be larger.")

# Assuming m is found and is an integer
plaintext = m.to_bytes((m.bit_length() + 7) // 8, 'big')
print("Decoded message:", plaintext.decode())
```


Then execute it to get a flag

```
n00bz{crypt0_1s_1nc0mpl3t3_w1th0ut_rs4!!}
```

Original writeup (https://github.com/ChickenLoner/Write-Up/blob/main/Unlisted%20Labs/n00bz%20CTF%202024%20Write-up%20(ByTheW4y%20Team).md).