The challenge itself hints towards the primes being posted online, implying the use of factordb.com. Visit the site and input n to get the prime factorization! From there, it’s just standard RSA decryption:
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from Crypto.Util.number import *
n = 23690620655271165329693230765997410033604713853187305472268813793031152348107488119317901392104240429826482611449247251262846508667797483465355228800439339041030982259847598574606272955688345490638311164838117491821117626835340577511562130640807587611523935604871183668968359720411023759980144229161581597397061850707647104033348795132205561234674677139395868595692235525931999596382758921793937149945229459379437008216713404350896206374483356969246476531491049930769999387038678280465689487577291475554699094024761030833540509263174840007922218340417888061099317752496279552046029470370474619439450870110783844218281 e = 65537 ct = 11420169733597912638453974310976296342840438772934899653944946284527921765463891354182152294616337665313108085636067061251485792996493148094827999964385583364992542843630846911864602981658349693548380259629884212903554470004231160866680745154066318419977485221228944716844036265911222656710479650139274719426252576406561307088938784324291655853920727176132853663822020880574204790442647169649094846806057218165102873847070323190392619997632103724159815363319643022552432448214770378596825200154298562513279104608157870845848578603703757405758227316242247843290673221718467366000253484278487854736033323783510299081405
# factordb.com p = 136883787266364340043941875346794871076915042034415471498906549087728253259343034107810407965879553240797103876807324140752463772912574744029721362424045513479264912763274224483253555686223222977433620164528749150128078791978059487880374953312009335263406691102746179899587617728126307533778214066506682031517 q = n//p assert p*q == n phi = (p-1)*(q-1) d = inverse(e, phi) print(long_to_bytes(pow(ct, d, n)))
