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

Answer: Option C

$$\begin{array}{rl}& \iff n^{65}-n\\ & =n\left(n^{64}-1\right)\\ & =n\left(n^{32}-1\right)\left(n^{32}+1\right)\\ & =n\left(n^{16}-1\right)\left(n^{16}+1\right)\left(n^{32}+1\right)\end{array}$$
  $$=n\left(n^8-1\right)\left(n^8+1\right)\left(n^{16}+1\right)$$     $$\left(n^{32}+1\right)$$
  $$=n\left(n^4-1\right)\left(n^4+1\right)\left(n^8+1\right)$$     $$\left(n^{16}+1\right)$$  $$\left(n^{32}+1\right)$$
  $$=n\left(n^2-1\right)\left(n^2+1\right)\left(n^4+1\right)$$     $$\left(n^8+1\right)$$  $$\left(n^{16}+1\right)$$  $$\left(n^{32}+1\right)$$
  $$=\left(n-1\right)n\left(n+1\right)\left(n^2+1\right)\left(n^4+1\right)$$       $$\left(n^8+1\right)$$  $$\left(n^{16}+1\right)$$  $$\left(n^{64}+1\right)$$  $$\left(n^{32}+1\right)$$
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