Question

If $n$ is a positive integer and ${(3\sqrt{3}+5)}^{2n+1}=\alpha +\beta$ where $\alpha$ is an integer and $0<\beta <1$, then

• A. $\alpha$ is an even integer
• B. ${(\alpha +\beta )}^2$ is divisible by $2^{2n+1}$
• C. The integer just below ${(3\sqrt{3}+5)}^{2n+1}$ divisible by $3$
• D. $\alpha$ is divisible by $10$

Answer 1

Answer: (A), (D)

$\alpha$ is an even integer
$\alpha$ is divisible by $10$

Consider ${(3\sqrt{3}+5)}^{2n+1}-{(3\sqrt{3}-5)}^{2n+1}=\alpha$
Solving the above expression, only even terms were left in the binomial expansion,
$2\times {(}^{2n+1}{C}_1{3\sqrt{3}}^{2n}5+{}^{2n+1}{C}_3{3\sqrt{3}}^{2n-2}{5}^3+\cdots +{}^{2n+1}{C}_{2n+1}{5}^{2n+1})=\alpha$
Take $2$ and $5$ common out of this expression, then it becomes:
$10\times (\dots )=\alpha$
${(3\sqrt{3}+5)}^{2n+1}=\alpha +{(3\sqrt{3}-5)}^{2n+1}$
$\beta ={(3\sqrt{3}-5)}^{2n+1}$
$\beta$ is always less than $1$.
Option A:
$\alpha$ is an even integer, i.e. $\alpha =2\times 5\times (\dots )$
Option B:
$(\alpha +\beta {)}^2$ is not divisible by $2^{2n+1},\beta$ contains a irrational number.
Option C:

Integer Just below ${(3\sqrt{3}+5)}^{2n+1}$ is not divisible by $3$.
Check for $n=1$, it does not satisfy.
Option D:
$\alpha$ is divisible by $10$, i.e. $\alpha =10\times (\dots )$
Hence, Option A, D.