Let R =(5 √5+11)31= I +f, where I is an integer and f is the fractional part of R, then R f is equal to -

Question

Let R =(5 √5+11)31= I +f, where I is an integer and f is the fractional part of R, then R f is equal to - (A) 231 (B) 331 (C) 262 (D) 1

Tardigrade Admin · Accepted Answer

Let R prime=(5 √5-11)31 Now R - R prime=(5 √5+11)31-(5 √5-11)31 ⇒ R - R prime= Integer ⇒ I +f- R prime= Integer ⇒ f- R prime is an Integer but -1< f- R prime<1 so f- R prime=0 ⇒ f= R prime so R . f= R ⋅ R prime=(5 √5+11)31(5 √5-11)31 =431=262

Q. Let $R = (55 + 11)^{31} = I + f$ , where $I$ is an integer and $f$ is the fractional part of $R$ , then $R f$ is equal to -

$2^{31}$

$3^{31}$

$2^{62}$

1

Q. Let R=(55​+11)31=I+f, where I is an integer and f is the fractional part of R, then Rf is equal to -

231

331

262

1

Let R′=(55​−11)31 Now R−R′=(55​+11)31−(55​−11)31 ⇒R−R′= Integer ⇒I+f−R′= Integer ⇒f−R′ is an Integer but −1<f−R′<1 so f−R′=0⇒f=R′ so R.f=R⋅R′=(55​+11)31(55​−11)31 =431=262

Q. Let $R = (55 + 11)^{31} = I + f$ , where $I$ is an integer and $f$ is the fractional part of $R$ , then $R f$ is equal to -

$2^{31}$

$3^{31}$

$2^{62}$