Rating: 5.0
# ångstromCTF 2020
**category: Crypto**
**Challenge Name: Wacko Images**
**Points: 90**
**Description:**
> 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
> The flag is actf{x#xx#xx_xx#xxx} where x represents any lowercase letter and # represents any one digit number.
We get an encrypted image and encryption script:
image-encryption.py :
```python
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')
```
enc.png :

**Solution:**
From the script we gather that for correct 'j' the ``` e = (j * 251 + pixel[i]) / key[i] ``` gives us the original value before encryption, so we loop through 'j' in range(0, max(key)) and for each 'e' we check if it's a floating point number or not, if not, it's possibly the right one. We add a few lines to the encryption script and hope we get an image that is close enough to original image.
image-decryption.py :
```python
from numpy import *
from PIL import Image
enc = Image.open(r"enc.png")
img = array(enc)
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 j in range(0,41):
e = (j * 251 + pixel[i]) / key[i]
eint = int(e)
if eint == e:
pixel[i] = e
break
img[x][y] = pixel
flag = Image.fromarray(img)
flag.save('flag.png')
```
flag.png :

Close enough.
Flag: actf{m0dd1ng_sk1llz}