| Keygenme #1 by VladMetz |
Each string in Flag.txt is a key
file = open('Flag.txt','w')
for a in range(33,126):
for b in range(33,126):
for c in range(33,126):
for d in range(33,126):
for e in range(33,126):
for f in range(33,126):
if ((e-f)*(e-f)+136*(c-b)*(d-a) == 1632):
temp=chr(a)
temp+='xx'
temp+=chr(b)
temp+='x'
temp+=chr(c)
temp+=chr(d)
temp+='xx'
temp+=chr(e)
temp+=chr(f)
file.writelines(temp)
file.write("\n")
file.close()
|
2022-07-22 11:13 |
| Cookie's Byte Playground |
[Click to reveal]I found clear password in x64DBG: DQN34KK6717O8799RQ61
|
2022-07-22 08:11 |
| Simple Crackme |
[Click to reveal]The Flag is your name's first five chars and 'hijkl'
Your name: randomchar
Password: randohijkl
|
2022-07-22 06:51 |
| CrackmeApp |
[Click to reveal]5up3rH1dd3nPa55w0rd
Nice challenge
|
2022-07-22 01:33 |