I ofc didn't solve this problem during the exam, but I did find a clue: we can construct a polynomial function with the outputs of the permutation as the coefficients of the polynomial, and -2 is a root of this function. So, (x+2) is a factor of this function. I spent some time trying to figure out any conclusions that could be drawn from this, but failed miserably. I did however also note that plugging '1' into the function will give the sum of the coefficients which is just the sum of 'n' natural numbers. This was where I came to dead end, so I ultimately left the question.

The next Tuesday, while discussing this question with my friends, I told Nimai and Jaswanth about my findings, and we tried to find some natural numbers that satisfied the conclusion to reverse engineer the solution. Then Nimai had the big brain idea to express f(x) as a product of (x+2) and some other function g(x). Now, plugging '1' allows us to use the earlier finding to our advantage and narrow down that the sum of n natural numbers must be divisble by 3. From here, it is quite clear to check the three possible cases of n, giving us our final answer. 

This solution is much more organic and linear compared to the given solution. Anyone could have stumbled upon this solution and gotten 17 marks for free. The other two questions were much more labour intensive, with complex geometrical constructions and an elaborate combinatorics set up. This question just requires some knowledge of functions and some big brain thinking.