But clearly not tested enough... can you get the flag?
nc 167.172.123.213 12345
Category: Cryptography
First let's look at otp.py:
from random import seed, randint as w
from time import time
j = int
u = j.to_bytes
s = 73
t = 479105856333166071017569
_ = 1952540788
s = 7696249
o = 6648417
m = 29113321535923570
e = 199504783476
_ = 7827278
r = 435778514803
a = 6645876
n = 157708668092092650711139
d = 2191175
o = 2191175
m = 7956567
_ = 6648417
m = 7696249
e = 465675318387
s = 4568741745925383538
s = 2191175
a = 1936287828
g = 1953393000
e = 29545
g = b"rgbCTF{REDACTED}"
if __name__ == "__main__":
seed(int(time()))
print(f"Here's 10 numbers for you: ")
for _ in range(10):
print(w(5, 10000))
b = bytearray([w(0, 255) for _ in range(40)])
n = int.from_bytes(bytearray([l ^ p for l, p in zip(g, b)]), 'little')
print("Here's another number I found: ", n)That creates a seed based on the current time, prints out 10 "random" numbers using that seed, then generates 40 more "random" numbers and uses them to xor the flag stored in g, and prints that out as a giant int.
kali@kali:~/Downloads$ python3 otp.py
Here's 10 numbers for you:
2112
8938
2679
9362
2191
2003
3016
9235
3833
2887
Here's another number I found: 183833425864737726945931152645483328541
time() returns the number of seconds since the epoch, which means we can measure time() locally before and after invoking otp.py and have a relatively small range of seeds to try. We'll know when we found the right seed because the 10 "random" numbers will match the values from otp.py.
#!/usr/bin/python3
from pwn import *
from time import time
from random import seed, randint
context.log_level='DEBUG'
start = int(time())
#p = process(['/usr/bin/python3', './otp.py'])
p = remote('167.172.123.213', 12345)
p.recvline() # Here's 10 numbers for you:
nums = []
for i in range(10):
line = p.recvline().rstrip()
nums.append(int(line))
p.recvuntil("Here's another number I found: ")
another_num = int(p.recvline().rstrip())
p.stream()
end = int(time())
print("start=%u end=%u" % (start, end))
print(nums)
print(another_num)
s = start
while s <= end:
seed(s)
found = True
for i in range(10):
r = randint(5, 10000)
if r != nums[i]:
found = False
if found:
print("FOUND SEED: %u" % s)
break
s += 1
g = another_num.to_bytes(40, 'little')
b = bytearray([randint(0, 255) for _ in range(40)])
n = bytearray([l ^ p for l, p in zip(g, b)]), 'little'
print(n)I tested this locally first, and then just had to run it against the remote server to get the flag:
kali@kali:~/Downloads$ ./otp_solve.py
[+] Opening connection to 167.172.123.213 on port 12345: Done
[DEBUG] Received 0xb1 bytes:
b"Here's 10 numbers for you: \n"
b'6370\n'
b'578\n'
b'948\n'
b'4674\n'
b'4804\n'
b'6605\n'
b'2529\n'
b'4644\n'
b'3823\n'
b'4330\n'
b"Here's another number I found: 14967871060053757142096566772157693944463496219343006011424058102584\n"
start=1594497862 end=1594497882
[6370, 578, 948, 4674, 4804, 6605, 2529, 4644, 3823, 4330]
14967871060053757142096566772157693944463496219343006011424058102584
FOUND SEED: 1594497882
(bytearray(b'rgbCTF{random_is_not_secure}\xfeF\xa9Q/O\x93/\xe2\x8d^\xf5'), 'little')
The flag is:
rgbCTF{random_is_not_secure}