Question

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

• A. 6
• B. 4
• C. 2
• D. 8

Answer 1

Answer: (B)

$(2+\sqrt{3}{)}^5={}^5{C}_0\cdot 2^5\cdot (\sqrt{3}{)}^0+{}^5{C}_1\cdot 2^4\cdot (\sqrt{3})$ ${}^5{C}_2\cdot 2^3(\sqrt{3}{)}^2+{}^5{C}_3\cdot 2^2\cdot (\sqrt{3}{)}^3+{}^5{C}_4\cdot 2\cdot (\sqrt{3}{)}^4+{}^5{C}_5\cdot 2^0(\sqrt{3}{)}^5$
$=32+80\sqrt{3}+240+120\sqrt{3}+90+9\sqrt{3}$ $\Rightarrow \left[(2+\sqrt{3}{)}^5\right]=723$
$\Rightarrow 723=3^1\times 241^1$
Now, positive integral divisors of
$723=(1+1)(1+1)=4$