2005 AMC 12B Problem 10

The sequence begins $2005, 133, 55, 250, 133, \ldots$ since $2^3 + 0^3 + 0^3 + 5^3 = 133,$ $1^3 + 3^3 + 3^3 = 55,$ $5^3 + 5^3 = 250,$ and $2^3 + 5^3 + 0^3 = 133.$ After the initial $2005,$ the terms cycle through $133, 55, 250$ with period $3.$ Term $n$ for $n \ge 2$ is the $((n-2)\bmod 3)$th entry of $133, 55, 250.$ Since $2005 - 2 = 2003 \equiv 2 \pmod 3,$ the $2005$th term is $250.$ Thus, the correct answer is E.