Let A={1,2,3, …, 10} and B={m/n: m, n ∈ A, m< n and gcd(m, n)=1}. Then n(B) is equal to :

Solution

Maths sir · 4 steps

Fix the denominator n, count the numerators coprime to it, then add up across n = 2 to 10.

1. Count by denominator. For a fixed n, the allowed m are the numbers from 1 to n-1 with gcd(m, n) = 1. There are exactly φ(n) of them. Here n = 1 gives nothing, because no m satisfies m < 1.
2. No repeats. Since gcd(m, n) = 1, every fraction is already in lowest terms. Two different reduced fractions can never be equal, so each pair (m, n) gives a different element of B.
3. List the numerators. n=2: 1. n=3: 1, 2. n=4: 1, 3. n=5: 1, 2, 3, 4. n=6: 1, 5. n=7: 1 to 6. n=8: 1, 3, 5, 7. n=9: 1, 2, 4, 5, 7, 8. n=10: 1, 3, 7, 9.

   n(B) = 1 + 2 + 2 + 4 + 2 + 6 + 4 + 6 + 4 = 31

4. So. B has 31 elements, which is option (b).

The idea to keep. To count reduced proper fractions with denominator up to N, add φ(n) for n = 2 to N. The gcd condition removes duplicates such as 1/2 and 2/4 for you.

Answer from the NTA final answer key. Solution by PrepWiser.