Question

Let $R={(5\sqrt{5}+11)}^{31}=1+f$, where $I$ is an integer and $f$ is the fractional part of $R$, then $R.f$ is equal to

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

Answer 1

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

$R=(5\sqrt{5}+11{)}^{31}=I+f$

$f\text{is the fractional part of}R$

$\text{means}f=R-\left[R\right]$

$\text{and Let}{R}^{\prime }=(5\sqrt{5}-11{)}^{31}$

$\text{Now}R-R^{\prime }=(5\sqrt{5}+11{)}^{31}-(5\sqrt{5}-11{)}^{31}$

${\Rightarrow }^{31}{C}_0\left(5\sqrt{5}{)}^0{+}^{31}{C}_1\right(5\sqrt{5}{)}^1+\dots {+}^{31}{C}_{31}(5\sqrt{5}{)}^{31}$

$-\left[{}^{31}{C}_0\right(5\sqrt{5}{)}^0{-}^{31}{C}_1(5\sqrt{5}{)}^1+\dots {-}^{31}{C}_{31}(5\sqrt{5}{)}^{31}]$

$\Rightarrow 2\text{( Sum of odd terms)}=\text{Even Integer}$

$\text{Now}R-R^{\prime }=1+f-R^{\prime }=\text{Even Integer}$

$f-R^{\prime }=\text{Integer}$

$\text{Now}0<f<1\text{and}0<R^{\prime }<1$

$\text{So}f-R^{\prime }=0\text{and}f=R^{\prime }$

$Rf=R{R}^{\prime }=(5\sqrt{5}+11{)}^{31}(5\sqrt{5}-11{)}^{31}$

$=4^{31}=2^{62}$