Crypto Caulingo
crypto
Well you’ve done it, you’re now an admin of the Cookie World Order. The clandestine organisation that seeks to control the world through a series of artfully placed tasty treats, bringing folks back in to their idea of what a utopian society would look like. Strangely enough, the webcam data is being fed to understand the properties of the entities you had originally seen. They seem to be speaking back into the camera (an unadvertised microphone) but it’s hard to understand what they want. You must- if nothing else ever was important in your life, you must make contact with these beautiful creatures! Also, what exactly is a “cauliflower”?
3d0b47fb89e15f694d2e48117886cf60cfdf4e3a554538a4e92771d551be2ab0
This is a RSA challenge with n, e and ciphertext given.
n:
17450892350509567071590987572582143158927907441748820483575144211411640241849663641180283816984167447652133133054833591585389505754635416604577584488321462013117163124742030681698693455489404696371546386866372290759608301392572928615767980244699473803730080008332364994345680261823712464595329369719516212105135055607592676087287980208987076052877442747436020751549591608244950255761481664468992126299001817410516694015560044888704699389291971764957871922598761298482950811618390145762835363357354812871474680543182075024126064364949000115542650091904557502192704672930197172086048687333172564520657739528469975770627
e:
65537
msg:
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
You can use http://factordb.com/ and try to lookup the factors of the modulus and if that is possible, the decryption is easy.
The factors are:
p = 115502906812186413716028212900548735990904256575141882752425616464266991765240920703188618324966988373216520827723741484031611192826120314542453727041306942082909556327966471790487878679927202639569020757238786152140574636623998668929044300958627146625246115304479897191050159379832505990011874114710868929959
q = 151086174643947302290817794140091756798645765602409645643205831091644137498519425104335688550286307690830177161800083588667379385673705979813357923016141205953591742544325170678167010991535747769057335224460619777264606691069942245683132083955765987513089646708001710658474178826337742596489996782669571549253
Solved the RSA task by using RsaCtfTool:
$ RsaCtfTool.py -n 17450892350509567071590987572582143158927907441748820483575144211411640241849663641180283816984167447652133133054833591585389505754635416604577584488321462013117163124742030681698693455489404696371546386866372290759608301392572928615767980244699473803730080008332364994345680261823712464595329369719516212105135055607592676087287980208987076052877442747436020751549591608244950255761481664468992126299001817410516694015560044888704699389291971764957871922598761298482950811618390145762835363357354812871474680543182075024126064364949000115542650091904557502192704672930197172086048687333172564520657739528469975770627 -e 65537 --uncipher 10222858773755433469764336996856768450760985527042400754296738815253035289209819346552764861826897178953694028617881565642777758738326213227107411789592929235575736410565920565861733942556636682795434764977287319815776058382523647988607790198049613544992176274587936815277174216903948943669207554417201538968298283788085816962846708118019145845233723263269669018766451198790115051407180739443164437436649867866601697336991972590981389609971438312980159261653258478120441047116705071636380222300805523491545451688999749093751323087018420829191731836502776412182336014125777600380527605582269800936591119274872274474047
flag: CTF{017d72f0b513e89830bccf5a36306ad944085a47}