Rating:

ASIS members conducted excellent crypto CTF.I will recommend everyone to solve it by themselves ,you will learn new concepts by solving them.[Link to the challenges](https://cryp.toc.tf/challenges)

Here are some of them I was able to solve it.

## roXen -:
> description:

Relationship with a cryptographer!
The Girlfriend: All you ever care about is crypto! I am sick of it! It's me or crypto!

The Cryptographer boyfriend: You meant to say it's you [XOR](roXen.py) cryptography.

The Girlfriend: I am leaving you.

22 solves

### Solution:

The given script:

python
#!/usr/bin/env python

from Crypto.Util.number import *
from secret import exp, flag, nbit

assert exp & (exp + 1) == 0

l = len(bin(x)[2:])
return (2 ** l - 1) ^ x

while True:
p = getPrime(nbit)
q = adlit(p) + 31337
if isPrime(q):
return p, q

p, q = genadlit(nbit)
e, n = exp, p * q

c = pow(bytes_to_long(flag), e, n)

print 'n =', hex(n)
print 'c =', hex(c)

Lets, start from the top:

So exp&(exp+1)==0 that just means exp must be a [Mersenne number](http://mathworld.wolfram.com/MersenneNumber.html). Then adlit(x) is returning the xor of passed number with the next biggest possible number with same length of bits which means basically that returned number contains reversed bits of passed number.

Then i realised that it's nothing difficult if you know the bits length you can just subtract the number from that max length bit number.
>For eg. adlit(44) + 44 = 63, adlit(120) +120 =127.
Simple isn't it

So we know n is 2048 bits long so p and q might be 1024 bits prime number.Let's check for it.
p+adlit(p)==2**1024-1
So we get two equations from here p+q=2**1024+31336 and p*q=n. Two equations two vars->quadratic p**2 -(2**1024 +31336)*p + n = 0. Sympy will do the work.Using solve() for quadratic function we get:

p = 91934396941118575436929554782758166784623142015203107928295225306949429527662253180027648166060067602233902389535868116051536080388999480377007211745229221564969130373120800620379012435790356909945473565305296926519232706950561924532325538399351352696805684504904629096892037592742285758390953849377910498739
q = 87834916545113015336000964296144306577174555879027549345134855850783246277838709952680829156347468418886211490335525241607253688425417142115840218894244902812798763051744684655923207165455737209507609386779708842318917975391900956941587572141475884466544826179681669143055208345737430546444402480246313669813

That was long thinking much left to travel ::wink::

Then, bruting public exponent

python
while i :
i+=1
e=2**i-1
if(gcd(e,h)==1):
d=invert(e,h)
if "CCTF" in (long_to_bytes(pow(c,d,n))):
print (long_to_bytes(pow(c,d,n)))
exit(0)

No result it went further around 100000 bits.
I was quite sure that it should be less than 2048 as then it will be greater than n.
Then I approached admin Factoreal for it. Then he told me what happen if gcd(e,phi) not equal to 1. So after tinkering around with my teammates I come to the point that's not possible LOL because we don't know but then I realised it that I can use CRT for it but it would become complicated maybe some simple approach will do then my teammate told me we can use precision for it. And we solved this after solving Clever girl question which boosted up our confidence for going with precision.

Note:We need to change the last char btw as precision was not too good enough for it.

So here was the final [script](script.py):

python
from Crypto.Util.number import *
import gmpy2
import pwn

gmpy2.get_context().precision=10000

l = len(bin(x)[2:])
return (2 ** l - 1) ^ x

p = 91934396941118575436929554782758166784623142015203107928295225306949429527662253180027648166060067602233902389535868116051536080388999480377007211745229221564969130373120800620379012435790356909945473565305296926519232706950561924532325538399351352696805684504904629096892037592742285758390953849377910498739
q = 87834916545113015336000964296144306577174555879027549345134855850783246277838709952680829156347468418886211490335525241607253688425417142115840218894244902812798763051744684655923207165455737209507609386779708842318917975391900956941587572141475884466544826179681669143055208345737430546444402480246313669813

assert isPrime(p)
assert isPrime(q)
assert adlit(p) + 31337 == q
assert p*q == n

h = (p-1)*(q-1)

for i in range(4096):
e = 2**i - 1
f = gmpy2.gcd(e,h)
try:
d = gmpy2.invert(e//f,h)
m = long_to_bytes(pow(gmpy2.mpz(pow(c,d,n)),1/f))
#print(pow(gmpy2.mpz(pow(c,d,n)),1/3))
if b"CCTF" in m:
print("i = ", i)
print(m)
except:
continue


>Flag:CCTF{it5_3a5y_l1k3_5uNd4y_MOrn1N9}

## Clever Girl -:
> description:

There is no barrier to stop a [clever girl](clever_girl.py)!

33 solves

### Solution:

The given script:
python
#!/usr/bin/env python

import gmpy2
from fractions import Fraction
from secret import p, q, s, X, Y
from flag import flag

assert gmpy2.is_prime(p) * gmpy2.is_prime(q) > 0
assert Fraction(p, p+1) + Fraction(q+1, q) == Fraction(2*s - X, s + Y)
print 'Fraction(p, p+1) + Fraction(q+1, q) = Fraction(2*s - %s, s + %s)' % (X, Y)

n = p * q
c = pow(int(flag.encode('hex'), 16), 0x20002, n)
print 'n =', n
print 'c =', c


So we were given n,c,x and y. This time there are three unknowns p,q and s. We now need to get three equations,first one is p*q=n and other two will get by equating numerator and denominator as they had lesser possiblity to have gcd() !=1 for which I did some maths for it. The next two are n+q=s+y and 2*n+p+q+1=2*s-x . So my teammate get the values for it quickly.

nonlinsolve([n+q - s - Y, 2*n+p+q+1-2*s+X, p*q-n], (p,q,s))

And he was so amazed to tell me that we got p and q from it.


(12604273285023995463340817959574344558787108098986028639834181397979984443923512555395852711753996829630650627741178073792454428457548575860120924352450409, 12774247264858490260286489817359549241755117653791190036750069541210299769639605520977166141575653832360695781409025914510310324035255606840902393222949771, 161010103536746712075112156042553283066813155993777943981946663919051986586388748662616958741697621238654724628406094469789970509959159343108847331259823138256432294313269203421659050140817247896562556361172161032623037006361409872307045649661542219054272855881029305328814299453639438693256941440232720246684)


Then same thing what we did for roxen as here gcd(e,phi) is 2 So we used precision for it. For lower precision the result was like this:
>b'CCTF{4L\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'

Then increasing it to 10000 we get the flag.

python
import gmpy2
from Crypto.Util.number import *

gmpy2.get_context().precision=10000

p = 12604273285023995463340817959574344558787108098986028639834181397979984443923512555395852711753996829630650627741178073792454428457548575860120924352450409
q = 12774247264858490260286489817359549241755117653791190036750069541210299769639605520977166141575653832360695781409025914510310324035255606840902393222949771
n = 161010103536746712075112156042553283066813155993777943981946663919051986586388748662616958741697621238654724628406094469789970509959159343108847331259823125490271091357244742345403096394500947202321339572876147277506789731024810289354756781901338337411136794489136638411531539112369520980466458615878975406339
c = 64166146958225113130966383399465462600516627646827654061505253681784027524205938322376396685421354659091159523153346321216052274404398431369574383580893610370389016662302880230566394277969479472339696624461863666891731292801506958051383432113998695237733732222591191217365300789670291769876292466495287189494
assert p*q == n

h = (p-1)*(q-1)
print(gmpy2.gcd(0x20002, h))
d = gmpy2.invert(0x20002//2, h)
print(long_to_bytes(gmpy2.sqrt(pow(c,d,n))))


>Flag: CCTF{4Ll___G1rL5___Are__T4len73E__:P}

## Time Capsule -:
> description:

You neither need 35 years nor even 20 years to [solve](time_capsule.py) this problem!

120 solves

### Solution:

The given script:
python
#!/usr/bin/python

from Crypto.Util.number import *
from secret import flag, n, t, z

def encrypt_time_capsule(msg, n, t, z):
m = bytes_to_long(msg)
l = pow(2, pow(2, t), n)
c = l ^ z ^ m
return (c, n, t, z)

print encrypt_time_capsule(flag, n, t, z)


So we have four values c,n,t and z.
This was most simple question, as we see z is unchanged. So I quickly xored the c with z then we have l^m for which we need to calcuate l only.

The value of l you can't calculate that fast.I wonder how they calculated it. Because pow(2,t) will be large enough as an exponent for 2. Then I think about fermat(literally), he concluded that pow(a,p-1,p)=1 if p is prime . But here it isn't The number n has 42 prime factors I get those factor from Elliptic Curve Method.

Then,I think why p-1 because its the euler totient of prime p. So, I can calculate totient of n in the same way easily.
Here's the final [script](capsule.py)
python
from Crypto.Util.number import *

c=30263951492003430418944035844723976843761515320480688994488846431636782360488888188067655841720110193942081554547272176290791213962513701884837856823209432209367951673301622535940395295826053396595886942990258678430777333636450042181585837395671842878310404080487115827773100028876775230121509570227303374672524063165714509957850966189605469484201028704363052317830254920108664916139026741331552127849056897534960886647382429202269846392809641322613341548025760209280611758326300214885296175538901366986310471066687700879304860668964595202268317011117634615297226602309205086105573924029744405559823548638486054634428
n=16801166465109052984956796702219479136700692152603640001472470493600002617002298302681832215942994746974878002533318970006820414971818787350153626339308150944829424332670924459749331062287393811934457789103209090873472485865328414154574392274611574654819495894137917800304580119452390318440601827273834522783696472257727329819952363099498446006266115011271978143149347765073211516486037823196033938908784720042927986421555211961923200006343296692217770693318701970436618066568854673260978968978974409802211538011638213976732286150311971354861300195440286582255769421094876667270445809991401456443444265323573485901383
t=6039738711082505929
z=13991757597132156574040593242062545731003627107933800388678432418251474177745394167528325524552592875014173967690166427876430087295180152485599151947856471802414472083299904768768434074446565880773029215057131908495627123103779932128807797869164409662146821626628200600678966223382354752280901657213357146668056525234446747959642220954294230018094612469738051942026463767172625588865125393400027831917763819584423585903587577154729283694206436985549513217882666427997109549686825235958909428605247221998366006018410026392446064720747424287400728961283471932279824049509228058334419865822774654587977497006575152095818

factors=(15013,583343756982313,585503197547927,609245815680559,612567235432583,634947980859229,635224892351513,639438000563939,654170414254271,654269804672441,667954470985657,706144068530309,721443717105973,737993471695639,744872496387077,746232585529679,795581973851653,815694637597057,817224718609627,841183196554507,864339847436159,873021823131881,884236929660113,899583643974479,922745965897867,942872831732189,951697329369323,971274523714349,1017566110290559,1018452110902339,1025985735184171,1027313536626551,1059774237802229,1067609726096989,1070689247726159,1079289330417443,1098516592571807,1107673252158281,1108654254305327,1110918654474373,1111516996694389,1112193819715441)
tot=1
for prime in factors:
tot*=prime-1
l=pow(2,pow(2,t,tot),n)
print(long_to_bytes(l^c^z))

Voila,
Flag:CCTF{_______________________________________________Happy_Birthday_LCS______________________________________________}
A nice flag

Original writeup (https://github.com/saurav3199/CTF-writeups/blob/master/CryptoCTF).