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If \([x]\) denotes the greatest integer function on \(x\), then the number of positive integral divisors of \(\left[(2+\sqrt{3})^5\right]\) is
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Verified Answer
The correct answer is:
4
\(\begin{aligned}
& (2+\sqrt{3})^5={ }^5 C_0 \cdot 2^5 \cdot(\sqrt{3})^0+{ }^5 C_1 \cdot 2^4 \cdot(\sqrt{3}) \\
& { }^5 C_2 \cdot 2^3(\sqrt{3})^2+{ }^5 C_3 \cdot 2^2 \cdot(\sqrt{3})^3 \\
& +{ }^5 C_4 \cdot 2 \cdot(\sqrt{3})^4+{ }^5 C_5 \cdot 2^0(\sqrt{3})^5 \\
& =32+80 \sqrt{3}+240+120 \sqrt{3}+90+9 \sqrt{3} \\
& (2+\sqrt{3})^5=362+209 \sqrt{3}=723.99 \\
& \Rightarrow\left[(2+\sqrt{3})^5\right]=723 \Rightarrow 723=3^1 \times 241^1 \\
\end{aligned}\)
Now, positive integral divisors of
\(723=(1+1)(1+1)=4\)
& (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} \\
& (2+\sqrt{3})^5=362+209 \sqrt{3}=723.99 \\
& \Rightarrow\left[(2+\sqrt{3})^5\right]=723 \Rightarrow 723=3^1 \times 241^1 \\
\end{aligned}\)
Now, positive integral divisors of
\(723=(1+1)(1+1)=4\)
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