See here : [https://berryberry.hatenablog.jp/](https://berryberry.hatenablog.jp/entry/2022/03/07/230635)

First of all, I got a python file "pass_validate.py".
## Given FIle
This is a program that judges whether input is correct or not.
```
def ValidatePassword(password):
valid = False
print("Attempting to validate password...")

if(len(password[::-2]) != 8):
print("Nah, you're not even close!!")
return False

pwlen = len(password)
chunk1 = 'key'.join([chr(0x98 - ord(password[c])) for c in range(0, int(pwlen / 2))])

if "".join([c for c in chunk1[::4]]) != '&e"3&Ew*':
print("Seems you're a terrible reverse engineer, come back after gaining some skills!")
return False

chunk2 = [ord(c) - 0x1F if ord(c) > 0x60 else (ord(c) + 0x1F if ord(c) > 0x40 else ord(c)) for c in password[int(pwlen / 2) : int(2 * pwlen / 2)]]

rand = [54, -45, 9, 25, -42, -25, 31, -79]
for i in range(0, len(chunk2)):
if(0 if i == len(chunk2) - 1 else chunk2[i + 1]) != chunk2[i] + rand[i]:
print("You're not a real hecker, try again! " + str(i))
return False

print("Password accepted!")
return True

print("\n************** Password Validator ***************")
print("Please enter password")

while True:
if ValidatePassword(input()):
exit()
else:
print("Try again!")
```
## Solution
In line 5, it determines if the input has 16 charactors. In line 11, it determines if difference from "0x98" for each characters in the input is equal to "&e"3&Ew*". So, we need to find characters which does not match the condition in line 19. It's easy to find them from back.
I solved with a paper and pen. Finally, got a FLAG below.
```
CTF{r3verS!ng_pyTh0n}
```