angstrom CTF 2020 write up

Shifter

Misc

 

What a strange challenge...

It'll be no problem for you, of course!

 

 

 

Solve 50 of these epic problems in a row to prove you are a master crypto man like Aplet123!
You'll be given a number n and also a plaintext p.
Caesar shift `p` with the nth Fibonacci number.
n < 50, p is completely uppercase and alphabetic, len(p) < 50
You have 60 seconds!
--------------------
Shift XECK by n=1

 

주어진 문자열을 n번째 피보나치 수열 만큼 shift 한 값을 보내면 된다.

 

 

EX

 

from pwn import *
 
def fibo(n):
    t = [0, 1]
 
    if n in (0, 1):
        f = n
    else:
        for i in range(2, n+1):
            f = sum(t)
            t = [t[1], f]
 
    return f
 
 
 
 
alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
 
p = remote("misc.2020.chall.actf.co", 20300)
 
for a in range(0,50):
    p.recvuntil("Shift ")
    taskstr = list((p.recvuntil(" ")).strip())
    p.recvuntil("=")
    n = fibo(int(p.recvline()))
 
 
    for i in range(len(taskstr)):
        taskstr[i] = alpha[(ord(taskstr[i]) - 65 + n) % 26]
 
    ans = ''.join(taskstr)
    print ans
    p.sendline(ans)
p.interactive()
 

 

 

 

flag : actf{h0p3_y0u_us3d_th3_f0rmu14-1985098}

---

Reasonably Strong Algorithm

Crypto

 

 

n = 126390312099294739294606157407778835887
e = 65537
c = 13612260682947644362892911986815626931

 

 

https://www.alpertron.com.ar/ECM.HTM

Integer factorization calculator

여기에서 n을 인수분해 해준다.

 

 

126 390312 099294 739294 606157 407778 835887 (39 digits) = 9 336949 138571 181619 × 13 536574 980062 068373

 

 

 

$ python rsatool.py -p 9336949138571181619 -q 13536574980062068373
Using (p, q) to initialise RSA instance

n = 126390312099294739294606157407778835887 (0x5f15e3803896f7b78f17d685eb590daf)

e = 65537 (0x10001)

d = 122116681453826726664714315157880092841 (0x5bded13dfcef588dae9e87c5c23eb8a9)

p = 9336949138571181619 (0x819381c5b26cbe33)

q = 13536574980062068373 (0xbbdb93397a0a3e95)

 

rsatool.py를 이용해서 d값을 구해준다.

 

c^d mod N

 

복호화

 

 

flag : actf{10minutes}

 

 

 

---

 

Wacko Images

Crypto

 

How to make hiding stuff a e s t h e t i c? And can you make it normal again? enc.png image-encryption.py

 

 

enc.png

 

from numpy import *
from PIL import Image
 
flag = Image.open(r"flag.png")
img = array(flag)
 
key = [41, 37, 23]
 
a, b, c = img.shape
 
for x in range (0, a):
    for y in range (0, b):
        pixel = img[x, y]
        for i in range(0,3):
            pixel[i] = pixel[i] * key[i] % 251
        img[x][y] = pixel
 
enc = Image.fromarray(img)
enc.save('enc.png')

각 픽셀의 rgb 값에 각각 41, 37, 23을 곱하고 251로 나눈 나머지 값을 저장하여 enc.png 이미지를 만든다.

 

 

역연산을 해주면 되는데 251로 나눈 몫을 모르기 때문에 t로 두고 브포를 때렸다.

 

 

ex.py :

from numpy import *
from PIL import Image
 
flag = Image.open(r"enc.png")
img = array(flag)
 
key = [41, 37, 23]
 
a, b, c = img.shape
 
for x in range (0, a):
    for y in range (0, b):
        pixel = img[x, y]
        for i in range(0,3):
            for t in range(0,41):
                if (251 * t + pixel[i]) % key[i] == 0:
                    pixel[i] = (251 * t + pixel[i]) / 23
        img[x][y] = pixel
 
enc = Image.fromarray(img)
enc.save('flag.png')