If [x] denotes the greatest integer function | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $x = (2+\sqrt{3})^5$ and $y = (2-\sqrt{3})^5$. Since $0 < 2-\sqrt{3} < 1$,we have $0 < y < 1$. Consider the expression $S = (2+\sqrt{3})^5 + (

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $x = (2+\sqrt{3})^5$ and $y = (2-\sqrt{3})^5$. Since $0 Consider the expression $S = (2+\sqrt{3})^5 + (2-\sqrt{3})^5$. Using the binomial expansion,the irrational terms cancel out,leaving $S$ as an even integer. $S = 2 	imes [^5C_0 \cdot 2^5 + ^5C_2 \cdot 2^3 \cdot 3 + ^5C_4 \cdot 2^1 \cdot 3^2] = 2 	imes [32 + 80 	imes 3 + 10 	imes 18] = 2 	imes [32 + 240 + 180] = 2 	imes 452 = 904$. Since $S = x + y = 904$ and $0 Thus,$[x] = 904 - 1 = 903$. Now,find the prime factorization of $903$: $903 = 3 	imes 301 = 3 	imes 7 	imes 43$. The number of positive integral divisors of $903 = 3^1 	imes 7^1 	imes 43^1$ is $(1+1)(1+1)(1+1) = 2 	imes 2 	imes 2 = 8$.

If $[x]$ denotes the greatest integer function of $x$,then the number of positive integral divisors of $\left[(2+\sqrt{3})^5\right]$ is

Similar Questions

The least positive integer greater than $1$ that divides $49^n + 16n - 1$ for all positive integers $n$ is

$\sqrt {\underbrace {111........1}_{200\,\text{digits}} - \underbrace {222.......2}_{100\,\text{digits}}} $ equals :-

The sum $1! + 4! + 7! + 10! + 12! + 13! + 15! + 16! + 17!$ is divisible by

The remainder when $7^n - 6n - 50$ $(n \in N)$ is divided by $36$ is:

If $2^{4n+3} + 3^{3n+1}$ is divisible by $P$ for all natural numbers $n$, then $P$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module