This is what we're given:

```
from sympy import *
from Crypto.Util.number import *
import random
p=getPrime(512)
q = nextprime(p + random.randint(10**9,10**10))
N=p*q
msg=b'UDCTF{REDACTED}'
pt = bytes_to_long(msg)
e = 65537
ct = pow(pt, e, N)
print(N)
print(e)
print(ct)
```

Seems very very similar to Grade 7. Only that some number between 10^9 and 10^10 was added to p. But... is that really enough to stop us from using the same attack?
After all, because we're squaring a every time, the time complexity of this operation reduces down to 10^5 worst case scenario. That should definitely run in time!

So just copy the code from Grade 7 to get the flag!

UDCTF{4n_RSA_5ch0ol_gr4dua73!!}

```
import gmpy2

n=150459385706485253914441877113384979120500190162060302508541299821944089329499694790524295291567135320851306118878915105907451588623958757693847782920309145753994837129247899050065917279292484317798035721308006529470560777407483024961882645653400385816416526996027114542480513056100444908809723540145733606413
e=65537
c=2307423154990120835718508986514267143655326830191633946685219656220840494132925634069678170936781595742873539412034460586639622885239343246714559828497111273868089182257159904851948098861145910137615097694560608874412798124055642460363270612990137075678106724613406247492210136960473648165963598137216228495

a = gmpy2.iroot(n, 2)[0] + 1
b = a*a - n
while not gmpy2.iroot(b, 2)[1]:
print(b)
b = a*a - n
a += 1

p = a - gmpy2.iroot(b, 2)[0]
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='')
```