Question

Let $R=(5\sqrt{5}+11{)}^{2n+1}$ and $f=R-\left[R\right]$ where $[.]$ denotes the greatest integer function, then the remainder when $R\times f$, for $n=2$, is divided by 15 is

Answer 1

Here, $f=R-\left[R\right]$ is the fractional part of $R$. Thus, if $I$ is the integral part of $R$, then $R=I+f=(5\sqrt{5}+11{)}^{2n+1}$, and $0<f<1$
Let $f^{\prime }=(5\sqrt{5}-11{)}^{2n+1}$. Then $0<f^{\prime }<1$ (as $5\sqrt{5}-11<1$ )
Now, $I+f-f^{\prime }=(5\sqrt{5}+11{)}^{2n+1}-(5\sqrt{5}-11{)}^{2n+1}$
$=2{[}^{2n+1}{C}_1(5\sqrt{5}{)}^{2n}\times 11{+}^{2n+1}{C}_3(5\sqrt{5}{)}^{2n-2}\times {11}^3+.....]$
$=$ an even integer

$\Rightarrow f-f^{\prime }$ must also be an integer
$\Rightarrow f-f^{\prime }=0\because 0<f<1,0<f^{\prime }<1$
$\Rightarrow f=f^{\prime }$
$\therefore ,Rf=R{f}^{\prime }=(5\sqrt{5}+11{)}^{2n+1}(5\sqrt{5}-11{)}^{2n+1}$
$=(125-121{)}^{2n+1}=4^{2n+1}=1024$