SHA2017 - Write up Crypto 200

Welcome to my blog.
Today, I will write up for SHA2017 CTF.

First, write for Crypto 200:
The challenge give me source code: source

If we want to get flag we must send to server signed of text start with ''ticket:admin|root|' :

if ticket.startswith('ticket:admin|root|'):
    self.request.sendall("Here you go!\n")
    self.request.sendall(flag + "\n")
    break
else:
    self.request.sendall("Sorry that function is only available to admin user root\n")

In the server provide 3 services:

Welcome to the secure login server, make your choice from the following options:
1. Register yourself as a user.
2. Collect flag
3. Sign a message
4. Exit

In the "Sign a message": We send to server one text in hex format, and server send to our sign of this text pading one more '\xff' in the beginning of text.

We know  $a^d*b^d  =  (a*b)^d$ then i had a good idea:

If we have some text start with '\xff' and multiple of they is 'ticket:admin|root|'+pading we can solve this challenge.

we have one code brute force to find some text start with '\xff' and multiple of they is 'ticket:admin|root|'+pading :

key = 'ticket:admin|root|'
found = False
i = 210
while not found:
    a = int((key + '\x00'*i).encode('hex'), 16)
    b = int((key + '\xff'*i).encode('hex'), 16)
    for j in range(200, i):
        if hex(int(a/pow(0xff, j)))[2:4] == 'ff' and hex(int(b/pow(0xff, j)))[2:4] == 'ff':
            print "i = %s and j = %s" %(i, j)
            rs = (int(a/pow(0xff, j))+int(b/pow(0xff, j)))/2
            print hex(rs)
            print pow(0xff, j)*rs > a
            print pow(0xff, j)*rs < b
            found = True
            break

i = 210 and j = 201
0xffa7878ccbd0bcba3a98d8cd75778e1d733a0af8af6aa0bbdd1058L
True
True

after run this code we have:

'ticket:admin|root|'+ pading = pow(0xff,201) * 0xffa7878ccbd0bcba3a98d8cd75778e1d733a0af8af6aa0bbdd1058L

We send null to server to get signed of ff and send a7878ccbd0bcba3a98d8cd75778e1d733a0af8af6aa0bbdd1058 to get  signed of it.

when we had two signed is number a and b we calculate (pow(a,201)*b)%n and send this to server and get the flag.

The challenge is finish!!!!