Present or Past

testing n= 1 gives an answer of 2009. Now I shall prove that the expression is divisible by 2009 for all n by using modular arithmetic, as follows... $2009 = 7^2 \times \ 41$ $\rightarrow\ 3100^n - 650^n -558^n +117^n \equiv 6^n - 6^n -5^n +5^n \equiv 0 mod(7)$ $\rightarrow\ 3100^n - 650^n -558^n +117^n \equiv 25^n - 35^n - 25^n +35^n \equiv 0 mod(41)$ $\therefore\ 2009| (3100^n - 650^n -558^n +117^n) \ as \ required$

based from the choices, the least value possible is 2...