Pertama, kita diberikan output.txt, isinya :
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n = 0x9ffa2a58ad286990fc5fe97b669e8cb2752e81fafa5ac774ea856d8ca124089ba4b06fe21a5d588c1dcb9602838d32cd70e50b85dec21fa79944543176c7a3b8b804ab754af2978f23b09f2905103dd5a4c748df8d9e9a079a5b38f6f69051b3c6582ebc2d2d199b3a97cb7e58af79b90fe08884626d188e194816bd51960a45 e = 0x3 c = 0x10652cdfaa6a6f6f688b98219cd32ce42c4d4df94afaea31cd94dfac50678b1f50f3ab1fd389f9998b6727ffd1a2c06ee6bde21ae85daef63fd0fa694a93f3674dc3f9ea0f2e3283a3d9897137aea12458aa3b8f96c61f3bf74a510bab7e7d8b7af52290d2621f1e06e52e6a7be4896c6465 |
Disitu kita tahu bahwa c diperoleh dari RSA. Karena nilai e sangat kecil, maka kemungkinan hasil m pangkat e akan lebih kecil dari nilai n. Maka kita bisa mendapatkan e dengan mengakar pangkat tiga nilai c. Berikut solver scriptnya :
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from gmpy2 import iroot from Crypto.Util.number import long_to_bytes as l2b n = 0x9ffa2a58ad286990fc5fe97b669e8cb2752e81fafa5ac774ea856d8ca124089ba4b06fe21a5d588c1dcb9602838d32cd70e50b85dec21fa79944543176c7a3b8b804ab754af2978f23b09f2905103dd5a4c748df8d9e9a079a5b38f6f69051b3c6582ebc2d2d199b3a97cb7e58af79b90fe08884626d188e194816bd51960a45 e = 0x3 c = 0x10652cdfaa6a6f6f688b98219cd32ce42c4d4df94afaea31cd94dfac50678b1f50f3ab1fd389f9998b6727ffd1a2c06ee6bde21ae85daef63fd0fa694a93f3674dc3f9ea0f2e3283a3d9897137aea12458aa3b8f96c61f3bf74a510bab7e7d8b7af52290d2621f1e06e52e6a7be4896c6465 m, _ = iroot(c,e) print(l2b(m)) |
Jalankan script diatas dan saya berhasil mendapatkan flag :
Flag : flag{080eaeb0d8f724bcb542562b3bb708e5}