Buuoj RSA4

题目

N = 331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004
c = 310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243
N = 302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114
c = 112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344
N = 332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323
c = 10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242

解题过程

1. 这里的数都是五进制。。。。先转10进制(^{[1]})
2. 然后通过猜测e等于3(^{[1]})
3. 有(m^{e} equiv c1 mod N1)
    (m^{e} equiv c2 mod N2)
    (m^{e} equiv c3 mod N3)
   中国剩余定理(^{[2]})求出(m^e)

from gmpy2 import *
from Crypto.Util.number import *
from functools import reduce

N1 = 331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004
c1 = 310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243

N2 = 302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114
c2 = 112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344

N3 = 332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323
c3 = 10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242

N = []
c = []
for i in range(3):
    # 将题目给的所有值转成十进制并放进列表中
    N.append(int(str(eval('N'+str(i+1))),5))
    c.append(int(str(eval('c'+str(i+1))),5))
    print('N'+str(i+1),'=',N[i])
    print('c'+str(i+1), '=', c[i])

def chinese_remainder(modulus, remainders):
    Sum = 0
    prod = reduce(lambda a, b: a*b, modulus)
    for m_i, r_i in zip(modulus, remainders):
        p = prod // m_i
        Sum += r_i * (inverse(p,m_i)*p)
    return Sum % prod
e = 3
# print(chinese_remainder(N,c))
pow_m_e = chinese_remainder(N,c)
# pow_m_e = 17446992834638639179129969961058029457462398677361658450137832328330435503838651797276948890990069700515669656391607670623897280684064423087023742140145529356863469816868212911716782075239982647322703714504545802436551322108638975695013439206776300941300053940942685511792851350404139366581130688518772175108412341696958930756520037
m = iroot(pow_m_e,3)[0]
print(long_to_bytes(m))

noxCTF{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}

参考

[1] https://blog.csdn.net/weixin_44017838/article/details/105058745
[2] https://blog.csdn.net/qq_39642801/article/details/104158699?utm_medium=distribute.pc_relevant.none-task-blog-baidujs-4