Let R be a relation from the set 1,2,3,,60 to itself such that R=(a,b) : b=pq, where p,q3 …

Question

Let $R$ be a relation from the set $\{1,2,3,\ldots,60\}$ to itself such that $R=\{(a,b) : b=pq, 	ext{ where } p,q\geq3 	ext{ are prime numbers}\}$. Then, the number of elements in $R$ is:

Accepted Answer

We have a set S = {1, 2, 3, ..., 60}, and a relation R defined on the set S. An element (a, b) belongs to the relation R if and only if b can be expressed as the product of two prime numbers p and q, where both p and q are greater than or equal to 3.

In terms of number theory, prime numbers are integers greater than 1 that have no divisors other than 1 and themselves. We are interested in prime numbers that are greater than or equal to 3, because p and q must both be greater than or equal to 3.

The primes greater than or equal to 3 and less than or equal to 60 are {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59}.

We need to find all possible values of b = p $$ \times $$ q such that b belongs to the set S. We start by choosing the smallest prime number (which is 3) and keep multiplying it with all the prime numbers until the product exceeds 60 :

1. If we start with p = 3, we can choose q to be 3, 5, 7, 11, 13, 17, or 19. This gives us 7 valid products that are less than or equal to 60.
2. If we start with p = 5, we can choose q to be 5, 7, or 11. This gives us 3 valid products that are less than or equal to 60.
3. If we start with p = 7, we can choose q to be 7. This gives us 1 valid product that is less than or equal to 60.

So, we have a total of 7 + 3 + 1 = 11 possible values for b = p $$ \times $$ q that satisfy the conditions.

Since a can be any number in the set S, there are 60 possible values for a for each of the 11 values of b. Therefore, the total number of elements in the relation R is 60 $$ \times $$ 11 = 660.

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