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 :

• A. 31
• B. 36
• C. 37
• D. 29

Hint:

To find the number of valid pairs, carefully compute the pairs where \( m <n \) and \( \gcd(m, n) = 1 \).

Answer 1

The Correct Option is A

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 \( \boxed{31} \).

Answer 2

Because each pair \((m,n)\) in the definition satisfies \(\gcd(m,n)=1\), the fraction \(\dfrac{m}{n}\) is in lowest terms. Two different coprime pairs \((m_1,n_1)\neq(m_2,n_2)\) cannot give the same reduced fraction, so each valid pair corresponds to a distinct element of \(B\).

For a fixed denominator \(n\) (with \(2\le n\le 10\)), the number of admissible numerators \(m\) (with \(1\le m1\), this equals the count of \(1\le m

So \[ n(B)=\sum_{n=2}^{10}\varphi(n). \] Compute \(\varphi(n)\) for \(2\le n\le 10\): \[ \begin{aligned} \varphi(2)&=1, &\varphi(3)&=2, &\varphi(4)&=2, &\varphi(5)&=4,\\ \varphi(6)&=2, &\varphi(7)&=6, &\varphi(8)&=4, &\varphi(9)&=6, &\varphi(10)&=4. \end{aligned} \]

Add them: \[ n(B)=1+2+2+4+2+6+4+6+4 = 31. \]

Answer

\(n(B)=31\). (Option 1)