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Mind Your Ps and Qs

In RSA, a small e value can be problematic, but what about N? Can you decrypt this? values

This is typical RSA decryption. We are given n, e and c.

If you don't know much about RSA, check out my overview!

All we need are the factors of N. Because it's small, we can try and check if the factors are known using FactorDB. And they are! So from here it's just standard RSA:

from Crypto.Util.number import inverse, long_to_bytes

c = 421345306292040663864066688931456845278496274597031632020995583473619804626233684
n = 631371953793368771804570727896887140714495090919073481680274581226742748040342637
e = 65537

p = 1461849912200000206276283741896701133693
q = 431899300006243611356963607089521499045809

phi = (p-1) * (q-1)
d = inverse(e, phi)
m = pow(c, d, n)

print(long_to_bytes(m))

# picoCTF{sma11_N_n0_g0od_55304594}

Mind Your Ps and Qs

In RSA, a small e value can be problematic, but what about N? Can you decrypt this? values

This is typical RSA decryption. We are given n, e and c.

If you don't know much about RSA, check out my overview!

All we need are the factors of N. Because it's small, we can try and check if the factors are known using FactorDB. And they are! So from here it's just standard RSA:

from Crypto.Util.number import inverse, long_to_bytes

c = 421345306292040663864066688931456845278496274597031632020995583473619804626233684
n = 631371953793368771804570727896887140714495090919073481680274581226742748040342637
e = 65537

p = 1461849912200000206276283741896701133693
q = 431899300006243611356963607089521499045809

phi = (p-1) * (q-1)
d = inverse(e, phi)
m = pow(c, d, n)

print(long_to_bytes(m))

# picoCTF{sma11_N_n0_g0od_55304594}