# CyberHeroines 2023
## Sophie Wilson
> [Sophie Mary Wilson](https://en.wikipedia.org/wiki/Sophie_Wilson) CBE FRS FREng DistFBCS (born Roger Wilson; June 1957) is an English computer scientist, who helped design the BBC Micro and ARM architecture. Wilson first designed a microcomputer during a break from studies at Selwyn College, Cambridge. She subsequently joined Acorn Computers and was instrumental in designing the BBC Micro, including the BBC BASIC programming language whose development she led for the next 15 years. She first began designing the ARM reduced instruction set computer (RISC) in 1983, which entered production two years later. - [Wikipedia Entry](https://en.wikipedia.org/wiki/Sophie_Wilson)> > Chal: Help this [designer of microprocessors](https://www.youtube.com/watch?v=R2SdSLCMKEA) solve this RSA challenge.>>  Author: [Prajakta](https://github.com/MeherP2246)>> `n = 784605825796844081743664431959835176263022075947576226438671818152943359270141637991489766023643446015742865872000712625430019936454136740701797771130286509865524144933694390307166660453460378136369217557779691427646557961148142476343174636983719280360074558519378409301540506901821748421856695675459425181027041415137193539255615283103443383731129040200129789041119575028910307276622636732661309395711116526188754319667121446052611898829881012810646321599196591757220306998192832374480348722019767057745155849389438587835412231637677550414009243002286940429895577714131959738234773350507989760061442329017775745849359050846635004038440930201719911010249665164009994722320760601629833907039218711773510746120996003955187137814259297909342016383387070174719845935624155702812544944516684331238915119709331429477385582329907357570479058128093340104405708989234237510349688389032334786183065686034574477807623401744101315114981390853183569062407956733111357740976841307293694669943756094245305426874297375074750689836099469106599572126616892447581026611947596122433260841436234316820067372162711310636028751984204768054655406327047223250327323182558843986421816373935439976256688835521454318161553726050385094844798296897844392636332777``e = 5``c = 268593521627440355433888284074970889184087304017829415653214811933857946727694253029979429970950656279149253529187901591829277689165827531120813402199222392031974802458605195286640398523506218117737453271031755512785665400604866722911900724895012035864819085755503886111445816515363877649988898269507252859237015154889693222457900543963979126889264480746852695168237115525211083264827612117674145414459016059712297731655462334276493`
Tags: _crypto_
## SolutionThis challenge is a `RSA` challenge and we are getting `n`, `e` and `c`. Since the exponent is small we can try a [`low public exponent attack`](https://crypto.stackexchange.com/questions/6713/low-public-exponent-attack-for-rsa). The idea is that the chance is there that no (modulo) rollover occured when the original text was encrypted. Leaving us with the option to just take the 5th rooth the retrieve the original message.
```pythonfrom gmpy2 import irootfrom Crypto.Util.number import long_to_bytesprint(long_to_bytes(iroot(c, e)[0]))```
Flag `chctf{d3516n3d_4c0rn_m1cr0_c0mpu73r}`