writeup

e Category: Cryptography Points: 144 Description: n = 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 e = 0b1101 c = 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 Writeup Here we are given a public exponent of 13, a ciphertext, and a modulus (both in hex). I don’t like looking at hex for RSA, so I converted it to decimal: n = 390489435257757050962694900628597750000579222652331667618284165302943905225288969762176855801097841580556025206135063119377780258773946551477397214530192584542559975339905571052158012999468964332437424628306511823269722453860362280496202533043021337796709339953817006833706504476313712943965322096302491056783227523784482002015145826987110574460307401934641874704050677704711801356268639773155081360828180903299964526589331336235306332567448562160526113866018006123967625576962689761521605387494709446518692462619862696168918816662536077370130661612070397479668829129920065326461187098803231397068051063929181738677526073658243869002572485447843169095810970975008795690074220899938136361892311637145808179133421447144085978655477817360358554032010516385983131407000011685574404126645116055363989608798299433959920227740092093920293556264000720381725702113290830326487540883170568741677111457568414739635182616494007311080902460736310863610217949372424386793484900002898788953599811705062201758177232830116056373978679467452122685972908266494259227709026880378075787599801726114638151543214172984729176207988155050429283715199503476101477015502266718187756254132831832109660502543721206835505701564820158859089751904108288138208040673 e = 13 c = 106043754914029053332380422656979154558759375897122425881860894698990092522305749145737374365974220063939891079764890878154070613914922418958583544498027475665662636594704078148882769433156473437121146773425968949033386468829167234570619323371997036112007508384568510674710537942785775371160931981856023034244395378977429902662365257811359886125721032794417995359223839236947456899351021253568576347533100384607077413225615309141290386668460232755119696828398399706666171431603664375206400701950963668907213087581791373592734473983283263803333104246685774546585505902951125395519774268152638482568171628156316175459200284173273491875899659385136852897094442953202427686882030389289946251863304968551409170000000000000 I noticed how small the e was and figured that it might be a possibility to take the 13th root of this ciphertext instead of finding the private exponent and decrypting from there....

pseudo-key Category: Crypto Points: 341 Description: Keys are not always as they seem… Note: Make sure to wrap the plaintext with flag{} before you submit! Author: Boolean Given: pseudo-key-output.txt && pseudo-key.py Writeup First thing I do here is look at the files given to us. Looks like we are given the ciphertext and some “pseudo-key”. Ciphertext: z_jjaoo_rljlhr_gauf_twv_shaqzb_ljtyut Pseudo-key: iigesssaemk From the program itself, it looks like our ciphertext is encrypted with some key, but we are given the key encrypted with itself....

uglybash Category: Misc Points: 359 Description: This bash script evaluates to echo dont just run it, dummy # flag{…} where the flag is in the comments. The comment won’t be visible if you just execute the script. How can you mess with bash to get the value right before it executes? Enjoy the intro misc chal. Author: arinerron Given: cmd.sh Writeup When looking at the file, we see a bunch of gibberish bash....