Let $A=\{1,2,3, \ldots, 10\}$ and $B=\left\{\frac{m}{n}: m, n \in A, m \lt n\right.$ and $\left.\operatorname{gcd}(m, n)=1\right\}$. Then $n(B)$ is equal to :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{A}=\{1,2, \ldots, 10\} \\ & \mathrm{B}\left\{\frac{\mathrm{m}}{\mathrm{n}}=\mathrm{m}, \mathrm{n} \in \mathrm{A}, \mathrm{m} < \mathrm{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1\right\} \\ & \mathrm{n}(\mathrm{B}) \end{aligned}$
$\mathrm{n}=2 \quad\left\{\frac{1}{2}\right\}$
$\mathrm{n}=3 \quad\left\{\frac{1}{3}, \frac{2}{3}\right\}$
$\begin{array}{ll}\mathrm{n}=4 & \left\{\frac{1}{4}, \frac{3}{4}\right\} \\ \mathrm{n}=5 & \left\{\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right\}\end{array}$
$\begin{array}{ll}\mathrm{n}=6 & \left\{\frac{1}{6}, \frac{5}{6}\right\} \\ \mathrm{n}=7 & \left\{\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\right\} \\ \mathrm{n}=8 & \left\{\frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}\right\}\end{array}$
$\begin{array}{ll}\mathrm{n}=9 & \left\{\frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9}\right\} \\ \mathrm{n}=10 & \left\{\frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10}\right\} \\ \mathrm{n}(\mathrm{B})=31\end{array}$ *

Let $A=\{1,2,3, \ldots, 10\}$ and $B=\left\{\frac{m}{n}: m, n \in A, m \lt n\right.$ and $\left…

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