Question

If $\left[x\right]$ denotes the greatest integer function of $x$, then $\left[\right(6\sqrt{6}+14{)}^{2n+1}]$ is

• A. an integer divisible by $2$
• B. an integer divisible by $3$
• C. an integer divisible by$5$
• D. an odd prime number

Answer 1

The correct option is A an integer divisible by $2$

Let ${(6\sqrt{6}+14)}^{2n+1}=I+F$
where $I$ is a natural number and $0<F<1$
Let $G={(6\sqrt{6}-14)}^{2n+1}$
$\Rightarrow 0<G<1$

Now,
$I+F-G={(6\sqrt{6}+14)}^{2n+1}-{(6\sqrt{6}-14)}^{2n+1}=2\left({}^{2n+1}{C}_1\right(6\sqrt{6}{)}^{2n}\cdot 14+{}^{2n+1}{C}_3(6\sqrt{6}{)}^{2n-2}\cdot {14}^3+\cdots )$

As $-1<F-G<1$, so
$F-G=0$
Therefore,
$I$ is an even integer.