For a real number $$x$$, let $$[x]$$ denote the largest integer $$ \leq x$$. For example, $$ [3,4] = 2 $$ and $$ [4,9] = 4 $$ Let $$N=[(\sqrt{27}+\sqrt{23})^6]$$. The remainder when $$N$$ is divided by $$1000$$ is
Solution
Let $$u=\sqrt{27}+\sqrt{23}$$ and $$v=\sqrt{27}-\sqrt{23}$$, so $$0<v<1$$. The conjugate sum $$u^6+v^6$$ is an integer, and using the recurrence with $$u^2+v^2=100$$ and $$u^2v^2=16$$ gives $$u^6+v^6=995200$$. Therefore, $$N=995200-1=995199$$, whose remainder modulo $$1000$$ is $$199$$.
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