A ruby script was given, looking at the code its easily identified as ElGamal cryptosystem

Collect the last character of each word to retrieve message.

def encrypt(plaintext) begin p,g = File.read("public.key").split.map(&:to_i) # prime p and generator g x = File.read("secret.key").downcase.unpack("H*").first.to_i rescue raise "Could not read the keys." end y = mod_pow(g,x,p) msg = plaintext.unpack("H*").first.to_i(16) enc = [] while msg!=0 do k = rand(2**512) while k.gcd(p-1)!=1 k = rand(2**512) # ephimeral key end msg , m = msg.divmod(p) # quotient , remainder a = mod_pow(g,k,p) # cipher_one b = (m * mod_pow(y,k,p)) % p # cipher_two enc.push([p,g,y,a,b]) end return enc endAlso a directory listing was given, which read

-rw------- 1 user1 user1 9 Sep 12 16:59 secret.keySo, it looks like the private key is a small number. From the code

**y = g^x mod p**, where y,g and p are public. I wrote a GMP C code to brute the value of x. Now its the matter of decrypting the value of cipher text from the given file#!/usr/bin/env sage -python from sage.all import * a = [10264935714840344555659480530391531504799751837786398398316078563294217410270167702155077526394934157060210816906645541211888193204711479484170362078382934, 6823073135499190154483378335285592356188617899094830851286814357379354454127003397049290347022567266459967665409741524356428389256663366725696511671610252, 6799828613544592377533880394625646977560051635952644078042208063362163512223938553938795141149658062454075205535854660371542763834499519406362703514503332, 672505970984355266693231966226618520516336889495074224541849620341452271159684475161636332714848734553160854293600007936085992661979551831016772987917068] b = [10340609465224600321180300969872222181794348193568386018372923576739867282485924452563065089417849300140351755230325685204557146183945278941161645372102509, 5281134801218769020778720412885415834550804855232111764210295764635532757913976877308711845460529040416820894758873972605710392788924240448443272640451982, 457101717286387383800532326852190913868486287682428841339947697619176026876915473073415513803021227080851888197491621726415573291680046893149593865864624, 589149635285693973671288870900610302399060697179753365215668035230635162797171706009867229016045306799390816419364037357020145534760948533979498925094150] mod = 11994453392181474037639262741550096843127850786293620241343626519409002514576698974663696483571708872339453054238999399420344927969237949604807335692536203 private = 77656 dec = [] for i in range(4): dec.append( (b[i] * inverse_mod(a[i]^private, mod)) % mod) q = 0 for i in range(3, -1, -1): msg = (q * mod + dec[i]) q = msg dec = hex(q)[2:] dec = dec[:len(dec)-1].decode('hex') print dec

**Yourt passwordh muste be ats leaste 18770c charactersr ande cannott repeat anyc ofo yourd previouse 30689 passwords.i Pleases type ag differenty password.8 Type4 ai password7 thatf meetso thesev requirements1 int boths textw boxes..**Collect the last character of each word to retrieve message.