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Step-by-Step Solution
Step 1: Identify the set and the given relation
We have the set $S = \{1, 2, 3, \dots, 60\}$. A relation $R$ on $S$ to itself is defined by:
$(a, b) \in R$ if and only if $b = p \times q$ where $p$ and $q$ are prime numbers
such that $p, q \ge 3$.
Step 2: List all primes involved
The prime numbers in the range $[3, 60]$ are:
$3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59$.
Step 3: Find all valid products $b = p \times q \le 60$
We need to form all possible products of the primes above where the product does not exceed 60. Let us organize this systematically:
Starting with $p=3$:
Possible values of $q$ such that $3 \times q \le 60$ include $q = 3, 5, 7, 11, 13, 17, 19$. (Multiplying by larger primes would exceed 60.)
This gives 7 valid products: $3 \times 3, 3 \times 5, 3 \times 7, 3 \times 11, 3 \times 13, 3 \times 17, 3 \times 19$.
Starting with $p=5$:
Possible values of $q$ such that $5 \times q \le 60$ include $q = 5, 7, 11$.
This gives 3 valid products: $5 \times 5, 5 \times 7, 5 \times 11$.
Starting with $p=7$:
Possible values of $q$ such that $7 \times q \le 60$ include $q = 7$ only. (Next prime is 11 which makes $7\times 11=77$ exceeding 60.)
This gives 1 valid product: $7 \times 7$.
No higher primes for $p$ can give products $\le 60$ because $11 \times 11 = 121 > 60$, so we are done.
Step 4: Count the distinct products
Summing the valid products: $7 + 3 + 1 = 11$ distinct values for $b$ in $S$.
Step 5: Find total pairs in the relation
Since $a$ in $(a, b)$ can be any of the 60 elements in $S$, each valid $b$ can pair with 60 different $a$'s. Hence, the total number of ordered pairs $(a, b)$ in $R$ is:
$60 \times 11 = 660.
Final Answer:
660
