Concatacollision

Concatacollision

Crypto

Challenge

DESCRIPTION To print the flag, looks like we need a 2FA token to verify. Although we have no token, our engineers have managed to secure the code of the checking server. I'm sure the check function isn't as secure as they think...

nc chals.whitehacks.ctf.sg 10001 Author: prokarius

ATTACHED FILES concat.py

Solution

workings

return int(str(int(str(x1) + s) % x0) + s) % x1 == 1


A2 + L*x0 = (x1 * (10**concat_exp)) + concat_numb
K*x1 + 1 = (A2 * (10**concat_exp)) + concat_numb


(K*x1 + 1 - concat_numb) / (10**concat_exp) + L*x0 = (x1 * (10**concat_exp)) + concat_numb
(K*x1 + 1 - concat_numb) / (10**concat_exp) = (x1 * (10**concat_exp)) + concat_numb - L*x0 
(K*x1 + 1 - concat_numb) = (x1 * (10**concat_exp))*(10**concat_exp) + concat_numb*(10**concat_exp) - L*x0*(10**concat_exp) 
(K*x1 + 1 - concat_numb) - concat_numb*(10**concat_exp) = (x1 * (10**concat_exp))*(10**concat_exp) - L*x0*(10**concat_exp) 

K*x1 + 1 - concat_numb*(10**concat_exp + 1) = (x1 * (10**concat_exp))*(10**concat_exp) - L*x0*(10**concat_exp) 

int(str(x1) + s) % x0) + s) % x1 == 1

Flag

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