n being any odd number greater than 1, n⁶⁵ - n is always divisible by :

Solution(By Examveda Team)

$$\eqalign{ & \Leftrightarrow {n^{65}} - n \cr & = n\left( {{n^{64}} - 1} \right) \cr & = n\left( {{n^{32}} - 1} \right)\left( {{n^{32}} + 1} \right) \cr & = n\left( {{n^{16}} - 1} \right)\left( {{n^{16}} + 1} \right)\left( {{n^{32}} + 1} \right) \cr} $$
  $$ = n\left( {{n^8} - 1} \right)\left( {{n^8} + 1} \right)\left( {{n^{16}} + 1} \right)$$     $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^4} - 1} \right)\left( {{n^4} + 1} \right)\left( {{n^8} + 1} \right)$$     $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = n\left( {{n^2} - 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$     $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
  $$ = \left( {n - 1} \right)n\left( {n + 1} \right)\left( {{n^2} + 1} \right)\left( {{n^4} + 1} \right)$$       $$\left( {{n^8} + 1} \right)$$  $$\left( {{n^{16}} + 1} \right)$$  $$\left( {{n^{64}} + 1} \right)$$  $$\left( {{n^{32}} + 1} \right)$$
Clearly, (n - 1), n and (n + 1) are three consecutive numbers and they have to be multiples of 2, 3 and 4 as n is odd.
Thus, the given number is definitely a multiple of 24