Question

Let $R=(5\sqrt{5}+11{)}^{31}=I+f$, where $I$ is an integer and $f$ is the fractional part of $R$, then $R\cdot f$ is equal to

• A. $2^{31}$
• B. $3^{31}$
• C. $2^{62}$
• D. $1$

Answer 1

The correct option is C $2^{62}$

Let $R^{\prime }={(5\sqrt{5}-11)}^{31}$
Now $R-R^{\prime }={(5\sqrt{5}+11)}^{31}-{(5\sqrt{5}-11)}^{31}$
$\Rightarrow R-R^{\prime }=$ $\Rightarrow i+f-R^{\prime }=$ = Integer
$\Rightarrow f-R^{\prime }$ is an integer but $-1<f-R^{\prime }<1$ So $F-R^{\prime }=0$ $\Rightarrow$ $f=R^{\prime }$
so $R.f=R.R^{\prime }$ $={(5\sqrt{5}+11)}^{31}{(5\sqrt{5}-11)}^{31}=4^{31}=2^{62}$