Update TimisoaraCTF

timisoaractf/Back_in_Time/README.md

@@ -1,13 +1,26 @@
-I always hated history class. I thought history would never come in handy.
+# Back in Time
 
-It contained two files:
-encrypt.py
-ciphertext.txt
+> I always hated history class. I thought history would never come in handy.
 
+> [encrypt.py](encrypt.py)
+> [ciphertext.txt](ciphertext.txt)
+
+------
 On reading the python code we can easily know that;
+
 Lowercase chars have been substituted by other Lowercase char.
+
 All other are where they are suppose to be.
 
+<br>
+
 We even know flag has timctf in start.
-So by simple reverse subsitution we get the flag as:
-timctf{Ps3udO_R4NdoM_1S_Not_RAnD0M}
+
+So, start by simple reverse subsitution;
+
+and making the words sensible in english;
+
+we get the flag
+
+------
+flag = timctf{Ps3udO_R4NdoM_1S_Not_RAnD0M}

timisoaractf/Those_Are_Rookie_Numbers/README.md

@@ -1,33 +1,50 @@
-You've overheard a discussion between two classmates arguing whether the size really matters.
-What does this have to do with RSA?
+# Not your average RSA
 
-Within this challenge there is single file given with RSA parameters n, e, c.
-We know, by RSA algorithm.
+> You've overheard a discussion between two classmates arguing whether the size really matters.
+> What does this have to do with RSA?
+
+> [params.txt](params.txt)
+------
+Within this challenge there is single file given with RSA parameters n, e, c
+
+We know, by RSA algorithm
+```
 m = c^d mod(n)
 c = m^e mod(n)
-We need to find m here.
+```
+We need to find m here
 
-So, we need to find d.
-For d we need phi(n).
+So, we need to find d
+
+For d we need *phi(n)*
+```
 phi(n) = (p-1)(q-1)
 where,
-
-p and q are prime numbers.
+p and q are prime numbers
 And, n = pq
-
-So, we go check factorizedb if we can factors.
-http://www.factordb.com/index.php?query=58900433780152059829684181006276669633073820320761216330291745734792546625247
-we get;
-p = 176773485669509339371361332756951225661
-q = 333197218785800427026869958933009188427
-
-So, now we get phi(n) = 58900433780152059829684181006276669632563849616305906563893514443102586211160
-
+```
+So, we go check [factordb](http://www.factordb.com/index.php?query=58900433780152059829684181006276669633073820320761216330291745734792546625247) if factors are available
+
+we get
+```python
+>>> p = 176773485669509339371361332756951225661
+>>> q = 333197218785800427026869958933009188427
+>>> phi_n = p * q
+>>> print(phi_n)
+58900433780152059829684181006276669632563849616305906563893514443102586211160
+```
 Now, we need to find d, which multiplicative inverse of e mod(phi(n)).
-Thus, we go on using a python code saved as d_find.py
-we get, d = 13699426463079754153895905688064380048051799131808763503874587494945034432713
 
-Now, for final c decode we use our c_decode.py
+Thus, we go on using [d_find.py](d_find.py)
+```Shell
+$ python d_find.py 58900433780152059829684181006276669632563849616305906563893514443102586211160 65537
+13699426463079754153895905688064380048051799131808763503874587494945034432713
+```
+Now, finally for flag [c_decode.py](c_decode.py)
+```Shell
+$ python c_decode.py 56191946659070299323432594589209132754159316947267240359739328886944131258862 13699426463079754153895905688064380048051799131808763503874587494945034432713 58900433780152059829684181006276669633073820320761216330291745734792546625247
+timctf{th0sE_rOoKIe_numB3rz}
+```
 
 And, voila the flag is:
 timctf{th0sE_rOoKIe_numB3rz}

timisoaractf/Those_Are_Rookie_Numbers/c_decode.py

@@ -1,6 +1,11 @@
-c = 56191946659070299323432594589209132754159316947267240359739328886944131258862
-d = 13699426463079754153895905688064380048051799131808763503874587494945034432713
-n = 58900433780152059829684181006276669633073820320761216330291745734792546625247
+import sys
+
+# c = 56191946659070299323432594589209132754159316947267240359739328886944131258862
+c = int(sys.argv[1])
+# d = 13699426463079754153895905688064380048051799131808763503874587494945034432713
+d = int(sys.arv[2])
+# n = 58900433780152059829684181006276669633073820320761216330291745734792546625247
+n = int(sys.argv[3])
 
 p = pow(c,d,n)
 size = len("{:02x}".format(n)) // 2

timisoaractf/Those_Are_Rookie_Numbers/d_find.py

@@ -1,3 +1,5 @@
+import sys
+
 def egcd(a, b):
     if a == 0:
         return (b, 0, 1)
@@ -12,7 +14,6 @@ def modinv(a, m):
     else:
         return x % m
 
-a = 65537
-m = 58900433780152059829684181006276669632563849616305906563893514443102586211160
-
-print(modinv(a,m))
+print(modinv(int(sys.argv[2]),int(sys.argv[1])))
+# print(modinv(65537, 58900433780152059829684181006276669632563849616305906563893514443102586211160)
+# 58900433780152059829684181006276669632563849616305906563893514443102586211160