RSA 2
문제
RSA2 80
c: 0x217c8bf9b45601267624c3b1ba89ae93d04c8fae32dc15496262f36f48d06c0dc9e178a77b77a33708dcbe1fcd55ea9eb636fe5684c2f0f08df3389f47b36a128636671eba300491c829ed1e252b1bb4dbb3b93bc46d98a10bb5d55347752052ab45e143fd46799be1d06ac3ff7e8b1eb181dfbba8dfac3910202fd0b9a25befe E : 266524484526673326121255015126836087453426858655909092116029065652649301962338744664679734617977550306567819672969837450223062478394149960243362563995235387971047857994699247277712682103161537347874310994510059329875060868679654080020041070975648626636209785889112656335054840517934593236597457100751820027783 N : 412460203584740978970185080155274765823237615982150661072746604041385717906706098256415230390148737678989448404730885157667896599397615737297545930957425615121654272472589331747646564634264520011009284080299605233265170506809736069720838542498970453928922703911186788239628906189362646418960560442406497717567
Author: TnMch
풀이
공개키의 e가 크면 우선 wieners-attack을 해본다.
#python3 wiener.py
[+] Found the continued fractions expansion convergents of e/N.
[+] Iterating over convergents; Testing correctness through factorization.
[+] ...
[+] Factored N! :) derived keypair components:
[+] ++ SHA1(e): bbf1a420da781634bb5bd02dcd4869f0b5c63d68
[+] ++ d : 27979163639208238097581493168255260980791785886427784936313524512033423912647
[+] ++ SHA1(d): d6d19e38c1db77d3cde21737ece5f1b89beaac0b
[+] ++ SHA1(N): fda4043b9bf88780fd9e9c04e739befe4760f143
[+] ++ p : 17161999324236539064317754140395642765702739382246098180731387908161141783840551460502918960425597679930951305598922767957529406215641102387965223811833367
[+] ++ SHA1(p): d90f4f6e042a3c4594159ab02ca43cb6c01e9e62
[+] ++ q : 24033342257638708824373735251516351694011297880096219691762845250861823504037151195236378659607369447498245214333469418187337738705973799517129078918392601
[+] ++ SHA1(q): 122b81e391070f0e8f42b7c0214a2c230fc1591f
[+] ++ SHA1(phiN): e6128a71e36de5313d69e80076e7136265f6faae바로 나와 주었다.
찾은 d와 주어진 N, c를 가지고 복호화 하였다.
Bugs_Bunny{Baby_Its_Cool_Lik3_school_haHAha}