# 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
```pythonx = 1a=0while 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} ```