STARTER

1.RSA Starter 1

Find the solution to 101^17 mod 22663

print(pow(101,17,22663)) #19906

2.RSA Starter 2

“Encrypt” the number 12 using the exponent e = 65537 and the primes p = 17 and q = 23. What number do you get as the ciphertext?

b = 12 e = 65537 p, q = 17, 23 N = p * q print(pow(b, e, N)) #301

3.RSA Starter 3

Given N = p*q and two primes:

p = 857504083339712752489993810777

q = 1029224947942998075080348647219

What is the totient of N?

p = 857504083339712752489993810777 q = 1029224947942998075080348647219 phi = ( p - 1) * ( q - 1 ) print(phi) #882564595536224140639625987657529300394956519977044270821168

4.RSA Starter 4

Given the two primes:

p = 857504083339712752489993810777

q = 1029224947942998075080348647219

and the exponent:

e = 65537

What is the private key d?

import gmpy2 p = 857504083339712752489993810777 q = 1029224947942998075080348647219 e=65537 phi=(p-1)*(q-1) d=gmpy2.invert(e,phi) print(d) # 121832886702415731577073962957377780195510499965398469843281

5.RSA Starter 5

Use the private key that you found for these parameters in the previous challenge to decrypt this ciphertext:

c = 77578995801157823671636298847186723593814843845525223303932

pow(c, d, n) #13371337

6.RSA Starter 6

Sign the flag crypto{Immut4ble_m3ssag1ng} using your private key and the SHA256 hash function

A nice application of RSA, but still nice and straightforward to implement:

from hashlib import sha256N=15216583654836731327639981224133918855895948374072384050848479908982286890731769486609085918857664046075375253168955058743185664390273058074450390236774324903305663479046566232967297765731625328029814055635316002591227570271271445226094919864475407884459980489638001092788574811554149774028950310695112688723853763743238753349782508121985338746755237819373178699343135091783992299561827389745132880022259873387524273298850340648779897909381979714026837172003953221052431217940632552930880000919436507245150726543040714721553361063311954285289857582079880295199632757829525723874753306371990452491305564061051059885803 d=11175901210643014262548222473449533091378848269490518850474399681690547281665059317155831692300453197335735728459259392366823302405685389586883670043744683993709123180805154631088513521456979317628012721881537154107239389466063136007337120599915456659758559300673444689263854921332185562706707573660658164991098457874495054854491474065039621922972671588299315846306069845169959451250821044417886630346229021305410340100401530146135418806544340908355106582089082980533651095594192031411679866134256418292249592135441145384466261279428795408721990564658703903787956958168449841491667690491585550160457893350536334242689flag="crypto{Immut4ble_m3ssag1ng}" flag=flag.encode("gb2312") h=sha256(flag) h=h.hexdigest()c=pow(int(h,16),d,N) print(hex(c)) #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

PRIMES PART 1

1.Factoring

Factorise the 150-bit number 510143758735509025530880200653196460532653147 into its two constituent primes. Give the smaller one as your answer.

Use factor DB,and get 19704762736204164635843 , 25889363174021185185929
So the answer is 19704762736204164635843

2.Inferius Prime

Here is my super-strong RSA implementation, because it’s 1600 bits strong it should be unbreakable… at least I think so!

import gmpy2 from Crypto.Util.number import inverse,long_to_bytes n = 742449129124467073921545687640895127535705902454369756401331 p = 752708788837165590355094155871 q = 986369682585281993933185289261 e = 3 ct = 39207274348578481322317340648475596807303160111338236677373 phi=(p-1)*(q-1) d=inverse(e,phi) m=pow(ct,d,n)print(long_to_bytes(m)) # crypto{N33d_b1g_pR1m35}

3.Monoprime

Why is everyone so obsessed with multiplying two primes for RSA. Why not just use one?

from Crypto.Util.number import inverse,long_to_bytesn = 171731371218065444125482536302245915415603318380280392385291836472299752747934607246477508507827284075763910264995326010251268493630501989810855418416643352631102434317900028697993224868629935657273062472544675693365930943308086634291936846505861203914449338007760990051788980485462592823446469606824421932591 e = 65537 ct = 161367550346730604451454756189028938964941280347662098798775466019463375610700074840105776873791605070092554650190486030367121011578171525759600774739890458414593857709994072516290998135846956596662071379067305011746842247628316996977338024343628757374524136260758515864509435302781735938531030576289086798942 phi=n-1 d=inverse(e,phi) print(long_to_bytes(pow(ct, d, n))) #crypto{0n3_pr1m3_41n7_pr1m3_l0l}

4.Square Eyes

It was taking forever to get a 2048 bit prime, so I just generated one and used it twice.

If you’re stuck, look again at the formula for Euler’s totient.

5.Manyprime

Using one prime factor was definitely a bad idea so I’ll try using over 30 instead.

If it’s taking forever to factorise, read up on factorisation algorithms and make sure you’re using one that’s optimised for this scenario.

from Crypto.Util.number import inverse,long_to_bytes import gmpy2 n = 580642391898843192929563856870897799650883152718761762932292482252152591279871421569162037190419036435041797739880389529593674485555792234900969402019055601781662044515999210032698275981631376651117318677368742867687180140048715627160641771118040372573575479330830092989800730105573700557717146251860588802509310534792310748898504394966263819959963273509119791037525504422606634640173277598774814099540555569257179715908642917355365791447508751401889724095964924513196281345665480688029639999472649549163147599540142367575413885729653166517595719991872223011969856259344396899748662101941230745601719730556631637 e = 65537 ct = 320721490534624434149993723527322977960556510750628354856260732098109692581338409999983376131354918370047625150454728718467998870322344980985635149656977787964380651868131740312053755501594999166365821315043312308622388016666802478485476059625888033017198083472976011719998333985531756978678758897472845358167730221506573817798467100023754709109274265835201757369829744113233607359526441007577850111228850004361838028842815813724076511058179239339760639518034583306154826603816927757236549096339501503316601078891287408682099750164720032975016814187899399273719181407940397071512493967454225665490162619270814464 factors=[13099895578757581201,16656402470578844539,12132158321859677597,14963354250199553339,12834461276877415051,14278240802299816541, 14178869592193599187, 11492065299277279799, 10336650220878499841, 12955403765595949597, 10638241655447339831, 11403460639036243901, 12973972336777979701, 14100640260554622013, 11665347949879312361, 11328768673634243077,9303850685953812323, 11530534813954192171, 15998365463074268941, 13572286589428162097, 11282698189561966721, 9389357739583927789, 15824122791679574573, 17281246625998849649, 15669758663523555763, 14523070016044624039, 17138336856793050757, 11473665579512371723,9282105380008121879, 16898740504023346457, 15364597561881860737, 17174065872156629921] phi=1 for x in factors:phi *= (x-1) d = gmpy2.invert(e, phi) print(long_to_bytes(pow(ct, d, n))) #crypto{700_m4ny_5m4ll_f4c70r5}

PUBLIC EXPONENT

1.Salty

Smallest exponent should be fastest, right?

from Crypto.Util.number import inverse, long_to_bytes n = 110581795715958566206600392161360212579669637391437097703685154237017351570464767725324182051199901920318211290404777259728923614917211291562555864753005179326101890427669819834642007924406862482343614488768256951616086287044725034412802176312273081322195866046098595306261781788276570920467840172004530873767 e = 1 ct = 44981230718212183604274785925793145442655465025264554046028251311164494127485print(long_to_bytes(ct)) #crypto{saltstack_fell_for_this!}

2.Modulus Inutilis

My primes should be more than large enough now!

from Crypto.Util.number import long_to_bytes from gmpy2 import iroot n = 17258212916191948536348548470938004244269544560039009244721959293554822498047075403658429865201816363311805874117705688359853941515579440852166618074161313773416434156467811969628473425365608002907061241714688204565170146117869742910273064909154666642642308154422770994836108669814632309362483307560217924183202838588431342622551598499747369771295105890359290073146330677383341121242366368309126850094371525078749496850520075015636716490087482193603562501577348571256210991732071282478547626856068209192987351212490642903450263288650415552403935705444809043563866466823492258216747445926536608548665086042098252335883 e = 3 ct = 243251053617903760309941844835411292373350655973075480264001352919865180151222189820473358411037759381328642957324889519192337152355302808400638052620580409813222660643570085177957print(long_to_bytes(iroot(ct,3)[0])) #crypto{N33d_m04R_p4dd1ng}

總結(jié)

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