Question:

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

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To find the number of valid pairs, carefully compute the pairs where \( m <n \) and \( \gcd(m, n) = 1 \).
Updated On: Mar 24, 2025
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The Correct Option is A

Solution and Explanation

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 \( \boxed{31} \).

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