Question 9

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 $$199$$.

Get AI Help

Other Cracku’s Useful Resources:

JEE Questions JEE Question Bank JEE Mains Previous Papers JEE Advanced Previous Papers JEE Study Material PDF JEE Mains Mock Test JEE Advanced Mock Test JEE Daily Target

Free JEE Questions

JEE Physics Questions JEE Maths Questions JEE Chemistry Questions

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

Get AI Help

Other Cracku’s Useful Resources:

Free JEE Questions

50,000+ JEE Students Trusted Our Score Calculator

Download resources