Key Concepts and Formulas

• Definition of a Relation: A relation $R$ from a set $A$ to a set $B$ is a subset of $A \times B$. The elements of $R$ are ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.
• Number of Elements in a Relation: If the choice of the first element $a$ and the second element $b$ in an ordered pair $(a, b) \in R$ are independent, and there are $N_a$ possible values for $a$ and $N_b$ possible values for $b$, then the total number of elements in $R$ is $|R| = N_a \times N_b$.
• Prime Numbers: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Step-by-Step Solution

The relation $R$ is defined from the set $S = \{1, 2, 3, \ldots, 60\}$ to itself. This means that for any ordered pair $(a, b) \in R$, we must have $a \in S$ and $b \in S$. The condition for an ordered pair $(a, b)$ to be in $R$ is $b = pq$, where $p$ and $q$ are prime numbers and $p, q \ge 3$.

Step 1: Determine the Number of Possible Values for 'a'

The first element $a$ of any ordered pair $(a, b)$ in $R$ must belong to the set $S = \{1, 2, 3, \ldots, 60\}$. The number of possible values for $a$ is the total number of elements in $S$. Number of choices for $a = |S| = 60$.

Step 2: Determine the Number of Possible Values for 'b'

The second element $b$ of any ordered pair $(a, b)$ in $R$ must satisfy two conditions:

1. $b \in S$, which means $1 \le b \le 60$.
2. $b = pq$, where $p$ and $q$ are prime numbers and $p, q \ge 3$.

First, let's list the prime numbers greater than or equal to 3: $\mathcal{P}_{\ge 3} = \{3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, \ldots\}$

We need to find all possible products $b = pq$ such that $p, q \in \mathcal{P}_{\ge 3}$ and $b \le 60$. To ensure we count distinct values of $b$ only once and avoid redundant calculations (e.g., $3 \times 5$ and $5 \times 3$), we will impose the condition $p \le q$.

Let's systematically find the possible values of $b$ by iterating through possible values of $p$:

• If $p = 3$: We need $b = 3q \le 60$, which implies $q \le \frac{60}{3} = 20$. Since $q$ must be a prime number and $q \ge p=3$, the possible values for $q$ are prime numbers in the range $[3, 20]$. These are $3, 5, 7, 11, 13, 17, 19$. The corresponding values of $b$ are: $3 \times 3 = 9$ $3 \times 5 = 15$ $3 \times 7 = 21$ $3 \times 11 = 33$ $3 \times 13 = 39$ $3 \times 17 = 51$ $3 \times 19 = 57$ These are 7 distinct values for $b$.

• If $p = 5$: We need $b = 5q \le 60$, which implies $q \le \frac{60}{5} = 12$. Since $q$ must be a prime number and $q \ge p=5$, the possible values for $q$ are prime numbers in the range $[5, 12]$. These are $5, 7, 11$. The corresponding values of $b$ are: $5 \times 5 = 25$ $5 \times 7 = 35$ $5 \times 11 = 55$ These are 3 distinct values for $b$.

• If $p = 7$: We need $b = 7q \le 60$, which implies $q \le \frac{60}{7} \approx 8.57$. Since $q$ must be a prime number and $q \ge p=7$, the only possible value for $q$ is $7$. The corresponding value of $b$ is: $7 \times 7 = 49$ This is 1 distinct value for $b$.

• If $p = 11$: We need $b = 11q \le 60$, which implies $q \le \frac{60}{11} \approx 5.45$. However, we also require $q \ge p=11$. There are no prime numbers $q$ that satisfy both $q \le 5.45$ and $q \ge 11$. Thus, no further values of $p$ will yield valid products. For instance, if $p=11$, the smallest possible product is $11 \times 11 = 121$, which is greater than 60.

The set of all distinct possible values for $b$ is the union of the values found: $\{9, 15, 21, 33, 39, 51, 57\} \cup \{25, 35, 55\} \cup \{49\}$. All these values are distinct and are within the set $S = \{1, 2, \ldots, 60\}$. The total number of distinct values for $b$ is $7 + 3 + 1 = 11$.

Step 3: Calculate the Total Number of Elements in R

The number of possible values for $a$ is 60. The number of possible values for $b$ is 11. Since the choice of $a$ is independent of the choice of $b$, the total number of elements in the relation $R$ is the product of the number of choices for $a$ and the number of choices for $b$. $|R| = (\text{Number of choices for } a) \times (\text{Number of choices for } b)$ $|R| = 60 \times 11 = 660$

Common Mistakes & Tips

• Excluding Prime 2: The condition $p, q \ge 3$ is crucial. Ensure that the prime number 2 is not used in forming the product $b$.
• Ensuring Distinctness of 'b': When calculating products $pq$, systematically enforce $p \le q$ to avoid counting the same composite number multiple times (e.g., $3 \times 5$ and $5 \times 3$ produce the same value of $b=15$).
• Upper Bound for 'b': Remember that $b$ must be an element of the set $\{1, 2, \ldots, 60\}$, so all calculated products $pq$ must not exceed 60.

Summary

The relation $R$ consists of ordered pairs $(a, b)$ where $a$ can be any of the 60 elements in the set $\{1, 2, \ldots, 60\}$. The second element $b$ must be a product of two prime numbers $p, q$ (both $\ge 3$) and must also be within the set $\{1, 2, \ldots, 60\}$. By systematically listing all such possible values of $b$, we found there are 11 distinct values for $b$. Since the choice of $a$ and $b$ are independent, the total number of elements in $R$ is the product of the number of choices for $a$ and $b$, which is $60 \times 11 = 660$.

The final answer is $\boxed{660}$.