# baby_RSA![Screenshot_20220429_104010](https://user-images.githubusercontent.com/75040566/166266192-1d64c920-fdf9-4552-8061-32b25595db30.png)
q1 : the server gave me```n = 185063385289091e = 65537ct = 55694727860139```
n is too small so we can use factordb to get p , q ```p = 16287317q = 11362423```
q2 : ```n = 12347448648776893189566532619036563562480525660820263086702822672522098528734929074125965819944710650985212740765506206272168385990942482859944690513847960438837162763090832356777572036466407019256875307199111496998817558617131931111458443194115000444562077096351483370235071231398774800846778854148388513900052482628156850399468110278874042653969807262517330200217180503876129837759281408118590665789085541384771890179428393845354798537531187091427794887705974473631938190765328475090487140432877479904438492650145220684654503591690354167485580941256159270461045792982254129409195041570137614437640695994787420615401e = 3ct = 26480272848384180570411447917437668635135597564435407928130220812155801611065536704781892656033726277516148813916446180796750368332515779970289682282804676030149428215146347671350240386440440048832713595112882403831539777582778645411270433913301224819057222081543727263602678819745693540865806160910293144052079393615890645460901858988883318691997438568705602949652125
```e is too small and we know that plaintext = e root (((n * k + ciphertext ) ))so we can use bruteforce ```python from gmpy2 import irootfor i in range(100000):    m , j  = iroot((n  * i  + ct ) , e )    if j :        print(m) ```    q3 :  ``` n = 73716843574352951236588243934248480698132624417496345851966682591574212605147077702330878612113346228865858282119350707786530411593816226713041914526177389179378211608498488931990100350808611832491922263600393791957710547676740875194991618016137166706547491415699492033815966939572751531498560879804859800461e = 65537ct = 41235449780929443384166147522020083683889749974005612902804345492422348056977414525436540920715243341085765115642708161070411267919671311047708984574587151797616127366342359285612956528102401417549506433811968845087980794462826978617650901152446775365531634707700598379517429824118845723086343262295388510539```but he gave another info , that is the q close to p . then we can use factorization 