This is what we're given:

```
from Crypto.Util.number import *
n=166045890368446099470756111654736772731460671003059151938763854196360081247044441029824134260263654537
e=65537
msg=bytes_to_long(b'UDCTF{REDACTED}')
ct=pow(msg,e,n)
print(n)
print(e)
print(ct)
```

n seems suspiciously small compared to the Grade 1... if you look it up in [factordb.com](http://), n's factors are already known!
Therefore the problem is now trivial -- reuse code from grade 1 to decrypt

UDCTF{pr1m3_f4ct0r_the1f!}

```
n=166045890368446099470756111654736772731460671003059151938763854196360081247044441029824134260263654537
e=65537
c=141927379986409920845194703499941262988061316706433242289353776802375074525295688904215113445883589653
p = 51700365364366863879483895851106199085813538441759

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='')
```