RSA

RSA Signature

image

The private keys are

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N = 15216583654836731327639981224133918855895948374072384050848479908982286890731769486609085918857664046075375253168955058743185664390273058074450390236774324903305663479046566232967297765731625328029814055635316002591227570271271445226094919864475407884459980489638001092788574811554149774028950310695112688723853763743238753349782508121985338746755237819373178699343135091783992299561827389745132880022259873387524273298850340648779897909381979714026837172003953221052431217940632552930880000919436507245150726543040714721553361063311954285289857582079880295199632757829525723874753306371990452491305564061051059885803
d = 11175901210643014262548222473449533091378848269490518850474399681690547281665059317155831692300453197335735728459259392366823302405685389586883670043744683993709123180805154631088513521456979317628012721881537154107239389466063136007337120599915456659758559300673444689263854921332185562706707573660658164991098457874495054854491474065039621922972671588299315846306069845169959451250821044417886630346229021305410340100401530146135418806544340908355106582089082980533651095594192031411679866134256418292249592135441145384466261279428795408721990564658703903787956958168449841491667690491585550160457893350536334242689

so I made a python code.. (i figured out myself thats why its not really optimized xD)

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from Crypto.Util.number import long_to_bytes,bytes_to_long
import hashlib

N = 15216583654836731327639981224133918855895948374072384050848479908982286890731769486609085918857664046075375253168955058743185664390273058074450390236774324903305663479046566232967297765731625328029814055635316002591227570271271445226094919864475407884459980489638001092788574811554149774028950310695112688723853763743238753349782508121985338746755237819373178699343135091783992299561827389745132880022259873387524273298850340648779897909381979714026837172003953221052431217940632552930880000919436507245150726543040714721553361063311954285289857582079880295199632757829525723874753306371990452491305564061051059885803
d = 11175901210643014262548222473449533091378848269490518850474399681690547281665059317155831692300453197335735728459259392366823302405685389586883670043744683993709123180805154631088513521456979317628012721881537154107239389466063136007337120599915456659758559300673444689263854921332185562706707573660658164991098457874495054854491474065039621922972671588299315846306069845169959451250821044417886630346229021305410340100401530146135418806544340908355106582089082980533651095594192031411679866134256418292249592135441145384466261279428795408721990564658703903787956958168449841491667690491585550160457893350536334242689


flag = b'crypto{Immut4ble_m3ssag1ng}'
hashed = hashlib.new('sha256')
hashed.update(flag)

h = hashed.hexdigest()
m = int(h, 16)
s = pow(m, d, N)

print(s)

Monoprime

image

output.txt

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n = 171731371218065444125482536302245915415603318380280392385291836472299752747934607246477508507827284075763910264995326010251268493630501989810855418416643352631102434317900028697993224868629935657273062472544675693365930943308086634291936846505861203914449338007760990051788980485462592823446469606824421932591                                                                  
e = 65537
ct = 161367550346730604451454756189028938964941280347662098798775466019463375610700074840105776873791605070092554650190486030367121011578171525759600774739890458414593857709994072516290998135846956596662071379067305011746842247628316996977338024343628757374524136260758515864509435302781735938531030576289086798942

after factoring in factordb, I found out that the prime factor is n = p*1 which means n = p and q = 1
so we can easily compute the d and decrypt the cipher text

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from Crypto.Util.number import long_to_bytes, inverse

n = 171731371218065444125482536302245915415603318380280392385291836472299752747934607246477508507827284075763910264995326010251268493630501989810855418416643352631102434317900028697993224868629935657273062472544675693365930943308086634291936846505861203914449338007760990051788980485462592823446469606824421932591
e = 65537
ct = 161367550346730604451454756189028938964941280347662098798775466019463375610700074840105776873791605070092554650190486030367121011578171525759600774739890458414593857709994072516290998135846956596662071379067305011746842247628316996977338024343628757374524136260758515864509435302781735938531030576289086798942

d = inverse(e, n-1)
m = pow(ct, d, n)
print(long_to_bytes(m).decode)

the output is: crypto{0n3_pr1m3_41n7_pr1m3_l0l}

Manyprime

image

output.txt

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n = 580642391898843192929563856870897799650883152718761762932292482252152591279871421569162037190419036435041797739880389529593674485555792234900969402019055601781662044515999210032698275981631376651117318677368742867687180140048715627160641771118040372573575479330830092989800730105573700557717146251860588802509310534792310748898504394966263819959963273509119791037525504422606634640173277598774814099540555569257179715908642917355365791447508751401889724095964924513196281345665480688029639999472649549163147599540142367575413885729653166517595719991872223011969856259344396899748662101941230745601719730556631637
e = 65537
ct = 320721490534624434149993723527322977960556510750628354856260732098109692581338409999983376131354918370047625150454728718467998870322344980985635149656977787964380651868131740312053755501594999166365821315043312308622388016666802478485476059625888033017198083472976011719998333985531756978678758897472845358167730221506573817798467100023754709109274265835201757369829744113233607359526441007577850111228850004361838028842815813724076511058179239339760639518034583306154826603816927757236549096339501503316601078891287408682099750164720032975016814187899399273719181407940397071512493967454225665490162619270814464

factoring n in factordb gives us 30 factors. put them into a list and calculate phi with for loop.

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from Crypto.Util.number import long_to_bytes

n = 580642391898843192929563856870897799650883152718761762932292482252152591279871421569162037190419036435041797739880389529593674485555792234900969402019055601781662044515999210032698275981631376651117318677368742867687180140048715627160641771118040372573575479330830092989800730105573700557717146251860588802509310534792310748898504394966263819959963273509119791037525504422606634640173277598774814099540555569257179715908642917355365791447508751401889724095964924513196281345665480688029639999472649549163147599540142367575413885729653166517595719991872223011969856259344396899748662101941230745601719730556631637
e = 65537
ct = 320721490534624434149993723527322977960556510750628354856260732098109692581338409999983376131354918370047625150454728718467998870322344980985635149656977787964380651868131740312053755501594999166365821315043312308622388016666802478485476059625888033017198083472976011719998333985531756978678758897472845358167730221506573817798467100023754709109274265835201757369829744113233607359526441007577850111228850004361838028842815813724076511058179239339760639518034583306154826603816927757236549096339501503316601078891287408682099750164720032975016814187899399273719181407940397071512493967454225665490162619270814464

factors = [9282105380008121879, 9303850685953812323, 9389357739583927789, 10336650220878499841, 10638241655447339831, 11282698189561966721, 11328768673634243077, 11403460639036243901, 11473665579512371723, 11492065299277279799, 11530534813954192171, 11665347949879312361, 12132158321859677597, 12834461276877415051, 12955403765595949597, 12973972336777979701, 13099895578757581201, 13572286589428162097, 14100640260554622013, 14178869592193599187, 14278240802299816541, 14523070016044624039, 14963354250199553339, 15364597561881860737, 15669758663523555763, 15824122791679574573, 15998365463074268941, 16656402470578844539, 16898740504023346457, 17138336856793050757, 17174065872156629921, 17281246625998849649]

phi = 1
for factor in factors:
    phi *= (factor-1)

d = pow(e, -1, phi)
m = pow(ct, d, n)

print(long_to_bytes(m).decode())

output: crypto{700_m4ny_5m4ll_f4c70r5}

Modulus Inutilis

image

because n is big compared to e and ct, find the eth root of the ct using gmpy2

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from Crypto.Util.number import long_to_bytes, inverse
from gmpy2 import iroot
n = 17258212916191948536348548470938004244269544560039009244721959293554822498047075403658429865201816363311805874117705688359853941515579440852166618074161313773416434156467811969628473425365608002907061241714688204565170146117869742910273064909154666642642308154422770994836108669814632309362483307560217924183202838588431342622551598499747369771295105890359290073146330677383341121242366368309126850094371525078749496850520075015636716490087482193603562501577348571256210991732071282478547626856068209192987351212490642903450263288650415552403935705444809043563866466823492258216747445926536608548665086042098252335883
e = 3
ct = 243251053617903760309941844835411292373350655973075480264001352919865180151222189820473358411037759381328642957324889519192337152355302808400638052620580409813222660643570085177957


print(long_to_bytes(iroot(ct,e)[0]))

flag: crypto{N33d_m04R_p4dd1ng}

Modular Arithmetic

Greatest Common Divisor

CryptoHack

The Greatest Common Divisor (GCD), sometimes known as the highest common factor, is the largest number which divides two positive integers (a, b).

For a = 12, b = 8 we can calculate the divisors of a: {1, 2, 3, 4, 6, 12} and the divisor of b: {1, 2, 4, 8}. Comparing these two we see that gcd(a,b) = 4

a = {1, 2, 3, 4}

GeeksForGeeks

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Basic Euclidean Algorithm for GCD:

The algorithm is based on the below facts.

  • If we subtract a smaller number from a larger one (we reduce a larger number), GCD doesn’t change. So if we keep subtracting repeatedly the larger of two, we end up with GCD.
  • Now instead of subtraction, if we divide the larger number, the algorithm stops when we find the remainder 0.
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def gcd(a,b):
    if a == 0:
        return b
    return gcd(b % a, a)

Extended GCD

CryptoHack

Let a and b be positive integers.

The extended Euclidean algorithm is an efficient way to find integers u, v such that a*u + b*v = gcd (a,b)

GeeksForGeek

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def gcdExtended(a, b):

    if a == 0:
        return b, 0, 1
    gcd, x1, y1 = gcdExtended(b % a, a)

    x = y1 - (b//a) * x1
    y = x1

    return gcd, x, y