Modulus 10

Let $a_n$ be defined by the recurrence relation $a_0=2,a_1=10,$ $a_n = 10a_{n-1} - 5a_{n-2} \ \ (n \ge 2).$ Let $\alpha = 5+2\sqrt{5}, \beta = 5-2\sqrt{5}.$

Then an easy induction shows that $a_n = \alpha^n + \beta^n,$ and another easy induction on the recurrence relation shows that $10|a_n$ for $n \ge 1$ (note that $a_2 = 90$ ).

Note that $\lfloor \alpha^n \rfloor = \lfloor a_n - \beta^n \rfloor = a_n + \lfloor -\beta^n \rfloor.$ Now $0 < \beta < 1,$ so $\lfloor -\beta^n \rfloor = -1$ for $n \ge 1.$

Putting this all together, we get $\lfloor \alpha^{999} \rfloor = a_{999} + \lfloor -\beta^n \rfloor = a_{999} - 1 \equiv 9 \pmod{10}.$