Let \( S = \{p_1, p_2, \dots, p_{10}\} \) be the set of the first ten prime numbers. Let \( A = S \cup P \), where \( P \) is the set of all possible products of distinct elements of \( S \). Then the number of all ordered pairs \( (x, y) \), where \( x \in S \), \( y \in A \), and \( x \) divides \( y \), is _________.
The set \( A = S \cup P \) consists of \( S \), the set of the first ten primes, and \( P \), the set of all possible products of distinct elements of \( S \). Thus, \( |A| = 2^{10} - 1 = 1023 \), since there are \( 2^{10} \) subsets of \( S \), excluding the empty subset.
For each \( x \in S \), \( x \) divides exactly half of the elements of \( A \), as for every product that doesn't contain \( x \), there is a corresponding one that does. Hence, for each \( x \in S \), there are 512 elements in \( A \) divisible by \( x \).
Since there are 10 elements in \( S \), the total number of ordered pairs \( (x, y) \) such that \( x \) divides \( y \) is:
\[ 512 \times 10 = 5120. \]
The correct answer is \( \boxed{5120} \).
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: