**Tags:** programming

Rating:

# Challenge description

Let x = 1

Let calculation = (x*(x+1)) + (2 *(x + 1))

Let reversed_calc = reversed number of calculation [for example if calculation = 123, reversed_calc will be 321]

If reversed_calc can be divided by 4 without reminder then answer = answer + reversed_calc

Repeat all the calculations until you have x = 543

The final answer will be the flag when x = 543

Flag Format : KCTF{answer_here}

Example Flag : KCTF{123}

**Author : NomanProdhan**

-----------------------------------------------------------

sol.py

```python

x = 1

a=0

while x!=544:

calc=(x*(x+1)) + (2 *(x + 1))

k=str(calc)

x =x +1

reversed_calc= k[::-1]

reversed_calc=int(reversed_calc)

if reversed_calc % 4 ==0:

# print(x)

a=a+reversed_calc

print(a)

```

Output of sol.py :

```

12252696

```

``` KCTF{12252696} ```

Original writeup (https://github.com/j3seer/KnightCTF-2022-WriteUps/tree/main/KnightCTF%202022/Programming/Reverse%20the%20answer).