Let R be a relation from the set {1,2,3, …, 60} to itself such that R={(a, b): b=p q, where p, q ≥ 3 are prime numbers. Then, the number of elements in R is :

Solution

Maths sir · 4 steps

Here a is completely free, so count the allowed values of b and multiply by 60.

1. What R restricts. The condition only talks about b. So a can be any of the 60 numbers, and each allowed b gives exactly 60 ordered pairs.
2. Primes to use. Primes ≥ 3 are odd primes. The ones small enough to matter are 3, 5, 7, 11, 13, 17, 19. Since p and q may be equal, squares like 3 × 3 are allowed.

   Smaller prime	Products pq ≤ 60	Count
   3	9, 15, 21, 33, 39, 51, 57	7
   5	25, 35, 55	3
   7	49	1
   11 or more	11 × 11 = 121 > 60, none	0

3. Count the pairs. There are 7 + 3 + 1 = 11 allowed values of b. Each one pairs with all 60 values of a, so n(R) = 11 × 60.
4. So. n(R) = 660, which is option (b).

The idea to keep. When a relation puts a condition on only one coordinate, the other one is free. Count the valid values of the restricted one, then multiply by the size of the free set.

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