RSA4(中国剩余定理)

题目

给出3组n和c

N = 331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004
c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
c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
c = 10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242

解题思路

根据加密公式，可以构造出同余方程组

m^e = c1 (mod n1)

m^e = c2 (mod n2)

m^e = c3 (mod n3)

而解这样的同余方程组就要用到中国剩余定理

中国剩余定理

下面给出证明：

分析一下这样构造的解x中每一项的系数

M_i的因数中包含除了m_i 其他所有模数，因此对其他所有模数取模都为0

而M_i^-1是M_i(mod m_i)的逆元，因此M_iM_i^-1 = 1 (mod m_i)，从而有a_iM_iM_i^-1 = a_i (mod m_i)

所以，对所有的i，x = a_i (mod m_i)都成立，x是该方程组的解，证明完毕

那么具体思路如下：

根据中国剩余定理求解m^e
题目没有给出加密指数e，但是在这种题型中e一般都比较小，我们直接遍历求解

但是题目还有一个坑，就是所给的n和c都是5进制数，不仔细观察可能会漏掉这一点

附上代码：

import gmpy2
import  binascii#利用中国剩余定理求解同余方程，aList：余数，mList：模数
def CRT(aList, mList):M = 1for i in mList:M = M * i   #计算M = ∏ mi#print(M)x = 0for i in range(len(mList)):Mi = M // mList[i]   #计算MiMi_inverse = gmpy2.invert(Mi, mList[i]) #计算Mi的逆元x += aList[i] * Mi * Mi_inverse #构造x各项x = x % Mreturn xif __name__ == "__main__":#========== n c ==========n1 = "331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004"c1 = "310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243"n2 = "302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114"c2 = "112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344"n3 = "332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323"c3 = "10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242"cList = [int(c1,5), int(c2,5), int(c3,5)]nList = [int(n1,5), int(n2,5), int(n3,5)]m_e = CRT(cList, nList) #计算m^efor e in range(1, 10):  #遍历e求解m, f = gmpy2.iroot(m_e, e) #m_e开e次根print("加密指数e = %d："%e)m = hex(m)[2:]if len(m)%2 == 1:m = m + '0' #binascii.unhexlify()参数长度必须为偶数，因此做一下处理flag = binascii.unhexlify(m)print(flag)

运行后发现，当e = 3时get flag！

noxCTF{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}

flag

flag{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}

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