[LINE CTF 2022] X Factor

You can check that signature is made by $sig = pow(M, d, N)$ without hashing by check $M = pow(sig, e, N)$.

 

for a given message $M_1, M_2, \dots, M_7$, you have to find $e_1, \dots, e_7$ such that $\prod_{1 \leq i \leq 7} M_i^{e_i} = \text{0x686178656c696f6e}$.

 

It seems hard but when you factorize $M_1, M_2, \dots, M_7$ and $\text{0x686178656c696f6e}$, you can fit $e_1, \dots, e_7$ by based on each primes. I fit it by hands,,

 

 

solver.py

from Crypto.Util.number import *
#from sage.all import *

e = 0x10001
N = 0xa9e7da28ebecf1f88efe012b8502122d70b167bdcfa11fd24429c23f27f55ee2cc3dcd7f337d0e630985152e114830423bfaf83f4f15d2d05826bf511c343c1b13bef744ff2232fb91416484be4e130a007a9b432225c5ead5a1faf02fa1b1b53d1adc6e62236c798f76695bb59f737d2701fe42f1fbf57385c29de12e79c5b3

plains = [0x945d86b04b2e7c7, 0x5de2, 0xa16b201cdd42ad70da249, 0x6d993121ed46b, 0x726fa7a7, 0x31e828d97a0874cff, 0x904a515]

target = 7521425229691318126
# 7521425229691318126 = 2 * 197 * 947 * 2098711 * 9605087

#ps = [2,197,947,2098711,9605087]
'''
for p in plains:
  print(factor(p))

print("-----------")
'''
x = plains[1] * plains[5] * plains[3] * plains[3] * plains[6] * plains[6] // plains[4] // plains[2] // plains[0]

s0 = 0x17bb21949d5a0f590c6126e26dc830b51d52b8d0eb4f2b69494a9f9a637edb1061bec153f0c1d9dd55b1ad0fd4d58c46e2df51d293cdaaf1f74d5eb2f230568304eebb327e30879163790f3f860ca2da53ee0c60c5e1b2c3964dbcf194c27697a830a88d53b6e0ae29c616e4f9826ec91f7d390fb42409593e1815dbe48f7ed4
s1 = 0x3ea73715787028b52796061fb887a7d36fb1ba1f9734e9fd6cb6188e087da5bfc26c4bfe1b4f0cbfa0d693d4ac0494efa58888e8415964c124f7ef293a8ee2bc403cad6e9a201cdd442c102b30009a3b63fa61cdd7b31ce9da03507901b49a654e4bb2b03979aea0fab3731d4e564c3c30c75aa1d079594723b60248d9bdde50
s2 = 0x9444e3fc71056d25489e5ce78c6c986c029f12b61f4f4b5cbd4a0ce6b999919d12c8872b8f2a8a7e91bd0f263a4ead8f2aa4f7e9fdb9096c2ea11f693f6aa73d6b9d5e351617d6f95849f9c73edabd6a6fde6cc2e4559e67b0e4a2ea8d6897b32675be6fc72a6172fd42a8a8e96adfc2b899015b73ff80d09c35909be0a6e13a
s3 = 0x2b7a1c4a1a9e9f9179ab7b05dd9e0089695f895864b52c73bfbc37af3008e5c187518b56b9e819cc2f9dfdffdfb86b7cc44222b66d3ea49db72c72eb50377c8e6eb6f6cbf62efab760e4a697cbfdcdc47d1adc183cc790d2e86490da0705717e5908ad1af85c58c9429e15ea7c83ccf7d86048571d50bd721e5b3a0912bed7c
s4 = 0xa7d5548d5e4339176a54ae1b3832d328e7c512be5252dabd05afa28cd92c7932b7d1c582dc26a0ce4f06b1e96814ee362ed475ddaf30dd37af0022441b36f08ec8c7c4135d6174167a43fa34f587abf806a4820e4f74708624518044f272e3e1215404e65b0219d42a706e5c295b9bf0ee8b7b7f9b6a75d76be64cf7c27dfaeb
s5 = 0x67832c41a913bcc79631780088784e46402a0a0820826e648d84f9cc14ac99f7d8c10cf48a6774388daabcc0546d4e1e8e345ee7fc60b249d95d953ad4d923ca3ac96492ba71c9085d40753cab256948d61aeee96e0fe6c9a0134b807734a32f26430b325df7b6c9f8ba445e7152c2bf86b4dfd4293a53a8d6f003bf8cf5dffd
s6 = 0x927a6ecd74bb7c7829741d290bc4a1fd844fa384ae3503b487ed51dbf9f79308bb11238f2ac389f8290e5bcebb0a4b9e09eda084f27add7b1995eeda57eb043deee72bfef97c3f90171b7b91785c2629ac9c31cbdcb25d081b8a1abc4d98c4a1fd9f074b583b5298b2b6cc38ca0832c2174c96f2c629afe74949d97918cbee4a

sig = s1 * s5 * s3 * s3 * s6 * s6 * pow(s4,-1,N) * pow(s2,-1,N) * pow(s0,-1,N) % N

print(pow(sig, e, N) == 0x686178656c696f6e)

print(hex(sig))

print('LINECTF{'+hex(sig)[-32:]+'}')

I made a dumb mistakes during ctf, I don't take a modulo operation while recover a signiture🤣

'CTF > Crypto' 카테고리의 다른 글

[LINE CTF 2022] lazy_stek  (0) 2022.03.27
[LINE CTF 2022] Forward-or  (0) 2022.03.27
[LINE CTF 2022] X Factor  (0) 2022.03.27
[LINE CTF 2022] ss-puzzle  (0) 2022.03.27
[zer0pts CTF 2022] ok  (0) 2022.03.22
[zer0pts CTF 2022] EDDH  (0) 2022.03.22
[zer0pts CTF 2022] CurveCrypto  (0) 2022.03.22
  Comments
댓글 쓰기