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Cryptographer We are provided with noob.txt which has it contents:- U1FEQEdeS1JDSEBEXlZDUEFYSG5ZQ29TVVBFRFlXRFxvUUJFTA== and noob.py which contents read #!/usr/bin/env python2 # I AM NOOB :) import string from hashlib import md5 from itertools import izip, cycle import base64 import time def xor(data, key): return bytearray(a^b for a, ...

N-95 Have you seen a QR code? Definitely. Have you seen a masked QR code? WTF! This is one of my favorite challs from the CTF So it appears that the QR code has been nicely painted upon. Using stegsolve, we realize that the colors are in every bit plane, so we can’t just read off the QR code easily. And almost no scanner would decod...

XKCD.com/2247 v2 Why v2? because organizers messed up in v1 by providing plaintext in the challenge and the flag read wrapmeintheflagformatandputunderscoresaroundmywords weird flag but ok. As the name suggests it is Hill Cipher We are provided with the ciphertext and key, we just have to decrypt the ciphertext using the key. The ciphertext s...

XORed I was given the following equations. Can you help me decode the flag? Key 1 = 5dcec311ab1a88ff66b69ef46d4aba1aee814fe00a4342055c146533 Key 1 ^ Key 3 = 9a13ea39f27a12000e083a860f1bd26e4a126e68965cc48bee3fa11b Key 2 ^ Key 3 ^ Key 5 = 557ce6335808f3b812ce31c7230ddea9fb32bbaeaf8f0d4a540b4f05 Key 1 ^ Key 4 ^ Key 5 = 7b33428eb14e4b54f2f4a3acae...

Unexpected We are given 3 N’s and 3 C’s and given that all three share primes i.e N1 = p*q N2 = q*r N3 = r*p This is a problem of classic RSA with a small twist. Now, we know factoring N into their factors is hard. However this can be don easily considering we have a common factor between pairs of N i.e q is a common factor of N1 and N2, r i...

Smol E N = 163741039289512913448211316444208415089696281156598707546239939060930005300801050041110593445808590019811244791595198691653105173667082682192119631702680644123546329907362913533410257711393278981293987091294252121612050351292239086354120710656815218407878832422193841935690159084860401941224426397820742950923 E = 3 C1 = 1105245397...

Randomization 2 Again, we are provided a binary rand2 which runs on the server. On connecting to the server, we’re provided with 2 numbers and then we’re asked to guess the next numbers several(10) times. Going ahead, let’s use Ghidra to decompile this binary, and then work our ass off to understand the terrible decompiled code. Just kidding...

Randomization 1 We are provided a binary rand1 which supposedly runs on the server. We have to figure out how would capture the flag Lets decompile the binary quickly in Ghidra. Seeing the output of main function we get undefined8 main(void) { uint uVar1; undefined8 uVar2; long in_FS_OFFSET; int local_1c; int local_18; int local...