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  1. # Sharif University CTF 2016: High-speed RSA Keygen [Part2]
  2.  
  3. def egcd(a, b):
  4. if (a == 0):
  5. return [b, 0, 1]
  6. else:
  7. g, y, x = egcd(b % a, a)
  8. return [g, x - (b // a) * y, y]
  9.  
  10. def modInv(a, m):
  11. g, x, y = egcd(a, m)
  12. if (g != 1):
  13. raise Exception("[-]No modular multiplicative inverse of %d under modulus %d" % (a, m))
  14. else:
  15. return x % m
  16.  
  17. def decrypt(p, q, e, c):
  18. n = p * q
  19. phi = (p - 1) * (q - 1)
  20. d = modInv(e, phi)
  21. m = pow(c, d, n)
  22. return m
  23.  
  24.  
  25. n = 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
  26. e = 65537
  27. c = 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
  28.  
  29. p = 144299940222848685214153733110344660304144248927927225494186904499495592842937728938865184611511914233674465357703431948804720019559293726714685845430627084912877192848598444717376108179511822902563959186098293320939635766298482099885173238182915991321932102606591240368970651488289284872552548559190434607447
  30.  
  31. q = n / p
  32.  
  33. print "p:", p
  34. print "q:", q
  35.  
  36. m = decrypt(p, q, e, c)
  37.  
  38. print hex(m)[2:-1].decode("hex")
  39.  
  40. # Result
  41. # > python decrypt.py
  42. # p: 144299940222848685214153733110344660304144248927927225494186904499495592842937728938865184611511914233674465357703431948804720019559293726714685845430627084912877192848598444717376108179511822902563959186098293320939635766298482099885173238182915991321932102606591240368970651488289284872552548559190434607447
  43. # q: 144245059483864997316184517203073096336312163518349447278779492969760750146161606776371569522895088103056082865212093431805166968588430946389831338526998726900084424430828957236538023320369476765118148928194401317313951462365588911048597017242401293395926609382786042879520305147467629849523719036035657146109
  44. # The flag is ................................................ e7531f1bb9c95c19e823065d3e6a86b6 .........
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