We are given \( A = \{1, 2, 3, \dots, 10\} \) and the set \( B = \left\{ \frac{m}{n} : m, n \in A, m <n \text{ and } \gcd(m, n) = 1 \right\} \).
To find \( n(B) \), we list the pairs \( (m, n) \) where \( m <n \) and \( \gcd(m, n) = 1 \).
We compute this for each \( n \in A \):
For \( n = 2 \), the valid pairs are \( \left( \frac{1}{2} \right) \).
For \( n = 3 \), the valid pairs are \( \left( \frac{1}{3}, \frac{2}{3} \right) \).
For \( n = 4 \), the valid pairs are \( \left( \frac{1}{4}, \frac{3}{4} \right) \).
For \( n = 5 \), the valid pairs are \( \left( \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \right) \).
For \( n = 6 \), the valid pairs are \( \left( \frac{1}{6}, \frac{5}{6} \right) \).
For \( n = 7 \), the valid pairs are \( \left( \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \right) \).
For \( n = 8 \), the valid pairs are \( \left( \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8} \right) \).
For \( n = 9 \), the valid pairs are \( \left( \frac{1}{9}, \frac{2}{9}, \frac{4}{9}, \frac{5}{9}, \frac{7}{9}, \frac{8}{9} \right) \).
For \( n = 10 \), the valid pairs are \( \left( \frac{1}{10}, \frac{3}{10}, \frac{7}{10}, \frac{9}{10} \right) \).
Counting all the valid pairs, we get \( n(B) = 31 \). Thus, the answer is \( \boxed{31} \).
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to: