# WEIRD ENCRYPTION> Points: 377
## Description> I made this weird encryption I hope you can crack it.>>[File](https://mega.nz/file/rt1GiIhb#YZzFsf07O-BVKugJSJoRQkazgs6I_pLMD_zISg6VGt0)>>[File](https://mega.nz/file/rt1GiIhb#YZzFsf07O-BVKugJSJoRQkazgs6I_pLMD_zISg6VGt0)
## SolutionThe easiest challenge. Just a little logic!
I wrote a script to get the index value of each characters of the ciphertext from the given `main_string` and store them in a list.
For each letter there existed a pair : **quotient** `c1` and **remainder** `c2` in the list respectively.
From the script it's clear the divisor in the case is 16 and what we need is the dividend **ASCII value of letter** and convert it to character.
So I apply a simple arithmetic : **dividend = divisor * quotient + remainder** to find ASCII values, convert them to character and finally get the text back.
```py#!/bin/env python3
with open("Encrypted", "r") as f:	ct = f.read().replace("\n", "")
main_string="c an u br ea k th is we ir d en cr yp ti on".split()
st = []flag = []ch = ""i = 0
# To make a list st containing index of characters from main_stringwhile i<len(ct):
	if (ct[i]=='c' and ct[i+1]!='r') or ct[i]=='u' or ct[i]=='k' or ct[i]=='d':		ch = main_string.index(ct[i])		i += 1	else:		ch = main_string.index(ct[i:i+2])		i += 2	st.append(ch)
# Calculating ASCII values and converting into characterfor i in range(0,len(st),2):	flag.append(chr(16*st[i]+st[i+1]))print(''.join(flag))```
## Flag> Hello. Your flag is DarkCTF{0k@y_7h15_71m3_Y0u_N33d_70_Br3@k_M3}.
