crypto

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....

Mix Category: Crypto Points: 330 Description: At the BASEment no. 65536, A man is irritated with SHIFT key in his KEYBOARD as it’s a sticky key, A kid is having chocolate icecream with a SPOON. Author: Whit3_D3vi1 Given: flag.txt && chall_encrypted.txt Writeup I opened flag.txt and got absolutely pwned :( If you opened this then you are a n00b From the hint, we can see that the uppercase letters are telling us something....

Chonky E Category: Crypto Points: 100 Description: E Note: P>Q Author: AC Given: ChonkyE.txt Writeup From the text file given, we see two different cryptosystems in play: RSA and Schmidt-Samoa. We are given the ciphertext encrypted by the Schmidt-Samoa cryptosystem and the n and e from the RSA cryptosystem. I have given links to the Wiki pages for both cryptosystems below. Schmidt Samoa encryption is very similar to RSA encryption....