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This is also a RSA challenge which is more interesting, given a RSA-704 bit. Factors are found in the wiki. I used openssl to finish, this challenge. root@Vijay:# openssl rsa -inform PEM -text -pubin -in publickey.pem -modulus Public-Key: (704 bit) Modulus: 00:e1:34:18:93:fe:6e:68:16:ce:c8:a9:70:a3:9c: 00:fa:54:7c:7d:a2:cd:ed:ab:0a:62:b9:1c:46:51: a8:3f:96:38:0b:cf:ae:e2:6f:7e:86:61:07:90:63: 89:42:1b:1e:68:d0:a1:7a:ad:c9:87:0b:98:58:e9: 56:28:6e:39:99:e9:8c:ec:98:81:53:4a:c7:72:ae: 78:f5:e8:ab:a1:e2:f8:d3:03:95:77:02:9d:87 Exponent: 65537 (0x10001) Modulus=E1341893FE6E6816CEC8A970A39C00FA547C7DA2CDEDAB0A62B91C4651A83F96380BCFAEE26F7E866107906389421B1E68D0A17AADC9870B9858E956286E3999E98CEC9881534AC772AE78F5E8ABA1E2F8D3039577029D87 writing RSA key -----BEGIN PUBLIC KEY----- MHQwDQYJKoZIhvcNAQEBBQADYwAwYAJZAOE0GJP+bmgWzsipcKOcAPpUfH2ize2r CmK5HEZRqD+WOAvPruJvfoZhB5BjiUIbHmjQoXqtyYcLmFjpVihuOZnpjOyYgVNK x3KuePXoq6Hi+NMDlXcCnYcCAwEAAQ== -----END PUBLIC KEY-

This challenge had made me mad, Some how I finally I managed to solve the challenge. This is a RSA Crypto, given a cipher and public key. This is RSA low public exponent attack, e=3. root@Vijay:# openssl rsa -pubin -in pub.pem -text -noout -modulus Public-Key: (4096 bit) Modulus: 00:d1:0b:a0:e9:cd:6d:d6:c3:89:5f:cd:f4:17:db: 21:e5:81:22:60:89:c6:c7:58:7f:c4:1b:3d:78:df: f5:2c:0f:8c:29:dc:6b:e9:fc:cf:31:68:32:e6:ff: 6f:f0:49:6e:9e:56:6e:cb:c1:31:06:4e:b8:47:5d: 6c:1b:c8:28:be:4a:f4:54:ad:62:cb:f0:d1:d2:cd: 5a:59:8a:24:1c:52:b1:6d:8e:e1:da:0c:a9:cc:56: 30:3c:d0:70:71:0e:6c:18:1f:2a:31:c6:88:7e:52: cf:14:bd:76:f6:25:80:a8:46:92:f8:81:98:a9:38: 49:0f:b2:de:19:41:b1:10:85:83:3d:ed:ca:16:67: 3f:4a:e5:4b:e6:0f:e0:da:66:24:a5:3d:b2:32:dc: a6:c5:88:7d:72:3c:77:39:c4:76:ef:30:60:19:a0: 57:f1:c6:be:37:a5:b8:20:d0:91:9a:cf:fd:18:63: d2:2c:6f:a7:30:fe:12:e8:15:35:9d:68:a4:ec:e1: c0:1e:f7:b0:ec:d9:59:91:b3:d9:71:d0:09:27:99: 5e:d6:6e:d

There are three files, one.txt, one.txt.enc and second.txt.enc. Challenge is to decrypt the second.txt.enc using the key. So we got a message + cipher, so we got a hint that operation done using XOR. So XORing (Message ^ Cipher) = Key. #!/usr/bin/env python import hashlib """ one.txt This sentence is encrypted using XOR cipher. """ plain_text = open('one.txt','r').read().strip() """ one.txt.enc LAcbGEUKHQEGDgsaHU8bGEUcFgwAEhUNHQtSHhYQFghSMyorWAwbGw0cCkE= """ cipher_text = open('one.txt.enc','r').read().decode("base64") print plain_text print '---------------------------------------------------------------------' print cipher_text print '---------------------------------------------------------------------' plain_text = [ord(i) for i in plain_text] cipher_text = [ord(i) for i in cipher_text] key = '' for i in range(len(plain_text)): c = ((plain

Question: Zgyzhsxrksvi dzh lirtrmzoob wvevolkvw uli gsv Svyivd ozmtfztv. Gsv pvb uli gsrh ovevo rh svyivd. Given Hint: The Atbash cipher is a very common, simple cipher. It was for the Hebrew alphabet, but modified here to work with the English alphabet. Basically, when encoded, an "A" becomes a "Z", "B" turns into "Y", etc. The Atbash cipher can be implemented as an Affine cipher by setting both "a" and "b" to 25. We used online  tool http://rumkin.com/tools/cipher/atbash.php  to decrypt the encoded message. This is your encoded/decoded text: Atbash   cipher was originally developed for the Hebrew language. The key for this level is hebrew.

The Berlekamp–Massey algorithm is an algorithm that will find the shortest linear feedback shift register for a given binary output sequence. The algorithm will also find the minimal polynomial of a linearly recurrent sequence in an arbitrary field. I have used sage for computing this program. # Berlekamp-Massey Algorithm #from __future__ import print_function s = [GF(2)(0), 0, 1, 0, 0, 0, 0, 0, 1, 1, 0] #input sequence n = len(s) C = [GF(2)(1)] B = [GF(2)(1)] temp = [] T = [] L = 0 N = 0 m = -1 print '----- n',n print '-----------------------------------------------------------------------' while N < n: temp = B d = s[N] for i in range(1,L+1): d = d + C[i]*s[N-i] print '----- d ',d if d == 1: T = C temp = [ 0 for i in range(int(N-m))] + temp if len(C) < len(temp): C = C + [0 for i in range(len(temp)-len(C))] else: temp = temp +

""" TRAIL and DIVISION METHOD with PRIME_SIEVE """ def primes_sieve(limit): a = [True] * limit # Initialize the primality list a[0] = a[1] = False for (i, isprime) in enumerate(a): if isprime: yield i for n in xrange(i*i, limit, i): # Mark factors non-prime a[n] = False def trial_division(n): """Return a list of the prime factors for a natural number.""" if n == 1: return [1] primes = primes_sieve(int(pow(n,0.5)) + 1) # Prime factor is always less than SQRT(n)+1 prime_factors = [] for p in primes: if p*p > n: break while n % p == 0: prime_factors.append(p) n //= p if n > 1: prime_factors.append(n) return prime_factors t = trial_division(600851475143) print t Fermat Theorem import math def gcd(a, b): while a != b: if a > b:

#Baby Step Giant Step DLP problem y = a**x mod n #Example 70 = 2**x mod 131 y = 70 a = 2 n = 131 s = floor(sqrt(n)) A = [] B = [] for r in range(0,s): value = y*(a^r) % n A.append(value) for t in range(1,s+1): value = a^(t*s) % n B.append(value) print A print B x1,x2 =0,0 for r in A: for t in B: if r == t: x1 = A.index(r) x2 = B.index(t) print x1,x2 break print 'the value of x is ', ((x2+1)*s - x1) % n # Answer

## using SAGE RSA n = 80646413 p = floor(sqrt(80646413)) ### Finding Factor p*q = n while(true): if (n%p == 0): print p break p=p+1 q = n / p print q #### phi(n) n1 = (p-1) * (q-1) print n1 e =5 d = e^-1 % n1 print d ### got private key ###Fast Exponentiation Algorithms def expo(c,d,n): if (d==1): return c%n if (d==2): return c*c % n if (d%2==0): return expo(expo(c,d/2,n),2,n) else: return c*expo(expo(c,(d-1)/2,n),2,n) cipher = [72895864,15633602,38820479,60303684,7458706,60299530,20682371,54642689,26066811,32615038,35349196,76400140,38820479,56463813,80491201,76400140,35349196,69567074,26066811,76400140,74270178,76127647,76127647,15633602,76400140,60303684,38820479,56463813,60303684,76400140,72844764,76127647,69302434,15633602,80491201,76400140,6809712,26066811,76400140,42498798,60299530,76127647,69302434,80491201,33234011] flag = '' for c in cipher: flag = flag+ chr(ex