Cryptography

RSA - 1

RSA1_Challenge

On this challenge we’re provided a file with the following contents:

Ever used RSA Encryption?

cyphertext = 10400286653072418349777706076384847966640064725838262071
n = 23519325203263800569051788832344215043304346715918641803
e = 71

Instead of solving this challenge mathematically, I prefered an easier aproach. I used a Python tool called RsaCtfTool.

Once I installed it I executed the following command:

python3 RsaCtfTool.py -n 23519325203263800569051788832344215043304346715918641803 -e 71 --uncipher 10400286653072418349777706076384847966640064725838262071

And I got the flag as output:

dsc{t00_much_m4th_8898}

RSA - 2

On this challenge we’re given another file with the following content:

e: 3
c: 2780321436921227845269766067805604547641764672251687438825498122989499386967784164108893743279610287605669769995594639683212592165536863280639528420328182048065518360606262307313806591343147104009274770408926901136562839153074067955850912830877064811031354484452546219065027914838811744269912371819665118277221
n: 23571113171923293137414347535961677173798389971011031071091131271311371391491511571631671731791811911931971992112232272292332392412512572632692712772812832933073113133173313373473493533593673733793833893974014094194214314334394434494574614634674794874914995035095215235415475575635695715775875935996016076136176196316416436476536596616736776836917017097197277337397437517577617697737877978098118218238278298398538578598638778818838879079119199299379419479539679719779839919971009101310191431936117404941729571877755575331917062752829306305198341421305376800954281557410379953262534149212590443063350628712530148541217933209759909975139820841212346188350112608680453894647472456216566674289561525527394398888860917887112180144144965154878409149321280697460295807024856510864232914981820173542223592901476958693572703687098161888680486757805443187028074386001621827485207065876653623459779938558845775617779542038109532989486603799040658192890612331485359615639748042902366550066934348195272617921683

In order to solve it we use the same tool than in the previous challenge:

python3 RsaCtfTool.py -n 23571113171923293137414347535961677173798389971011031071091131271311371391491511571631671731791811911931971992112232272292332392412512572632692712772812832933073113133173313373473493533593673733793833893974014094194214314334394434494574614634674794874914995035095215235415475575635695715775875935996016076136176196316416436476536596616736776836917017097197277337397437517577617697737877978098118218238278298398538578598638778818838879079119199299379419479539679719779839919971009101310191431936117404941729571877755575331917062752829306305198341421305376800954281557410379953262534149212590443063350628712530148541217933209759909975139820841212346188350112608680453894647472456216566674289561525527394398888860917887112180144144965154878409149321280697460295807024856510864232914981820173542223592901476958693572703687098161888680486757805443187028074386001621827485207065876653623459779938558845775617779542038109532989486603799040658192890612331485359615639748042902366550066934348195272617921683 -e 3 --uncipher 278032143692122784526976606780560
4547641764672251687438825498122989499386967784164108893743279610287605669769995594639683212592165536863280639528420328182048065518360606262307313806591343147104009274770408926901136562839153074067955850912830877064811031354484452546219065027914838811744269912371819665118277221

This time it will take some time depending on the computer but it will finally output the flag:

dsc{t0-m355-w1th-m4th-t4k35-4-l0t-0f-sp1n3}

Forensics

Pirates

Pirates_Challenge

On this challenge we’re given a pcap file. We open it in wireshark and we find the following request:

Pirates1

The flag can be found by right-clicking and selecting the “Follow TCP Stream” option.

Pirates1 

dsc{H3_1S_th3_83sT_p1r4t3_1_H4V3_3V3r_s33n}