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
$723=(1+1)(1+1)=4$