This is what we're given:

```
from Crypto.Util.number import *
p=getPrime(512)
q=getPrime(512)
n=p*q
e=65537
msg=bytes_to_long(b'UDCTF{REDACTED}')
ct=pow(msg,e,n)
print(p)
print(n)
print(e)
print(ct)
```

We are given p, so finding q and therefore decrypting RSA is trivial! Here's a short demonstration of why for beginners:

Modulus = N = p * q ==> q = N / p
Euler's totient function = Φ(n) = (p - 1)(q - 1)
Private exponent = d = the modular inverse of e (mod Φ(n))
Message = c ^ d (mod N)

If you want to learn more about RSA check [https://bitsdeep.com/posts/attacking-rsa-for-fun-and-ctf-points-part-1/](http://) out!

UDCTF{y3a_b0i_b4by_RSA!}

```
p=7009789528005665925389589645247771843738610365138497450285434114825324963561464592190523618045678504210355855286077875965585945664408796837853566415684077
n=73061872549912499570368059392056653520123131891860048946474996807859190776947568485365189613739847632132597352816568237525325622321891749472819811314630053648031678826291232292672975634200777457699671848298242827252269004463672931479153540235625891818449660268924228002522141737330313259535617652381070426543
e=65537
c=8099012654842320180974620472267007973324910863630262955526926760464542904631823196320598910081443799605804614201671967967929893760527002416689993003801924422327762868245291561376910828637706061326005113536536357969201659290874169593264337355365186414719656091960977568710047843815328537885731546232759484717

q = n // p
phi = (p - 1) * (q - 1)
d = pow(e, -1, phi)
m = pow(c, d, n)
m = format(m, 'x')
for i in range(0, len(m), 2):
print(chr(int(m[i:i+2], 16)), end='')
```