Question ID: #386
Let $S=\{p_{1},p_{2}……,p_{10}\}$ be the set of 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), $x\in S, y\in A$ such that x divides y, is
Solution:
Set $S$ contains 10 distinct primes $\{p_1, \dots, p_{10}\}$.
Set $P$ contains all products of distinct elements of $S$.
Elements of $P$ are of the form $p_{i_1} p_{i_2} \dots p_{i_k}$ where $k \ge 2$.
Set $A = S \cup P$. Effectively, $A$ contains all products of distinct primes from $S$ taken 1 at a time (which is $S$), 2 at a time, …, up to 10 at a time.
Total elements in $A = 2^{10} – 1$ (excluding the empty product 1).
We need pairs $(x, y)$ such that $x \in S$, $y \in A$ and $x|y$.
For a fixed $x = p_i \in S$:
$y$ must be a multiple of $p_i$ in $A$.
$y$ is a product of distinct primes from $S$. For $p_i$ to divide $y$, $p_i$ must be one of the factors in the product formation of $y$.
Let’s choose the other factors of $y$ from the remaining 9 primes $S \setminus \{p_i\}$.
The remaining factors can be any subset of the remaining 9 primes.
Number of subsets of 9 elements = $2^9$.
Each subset combined with $p_i$ forms a valid $y \in A$.
(Note: The empty subset corresponds to $y = p_i$, which is in $S \subset A$. This is valid).
So, for a fixed $x$, there are $2^9$ choices for $y$.
Since there are 10 choices for $x$ (any of the 10 primes):
Total pairs = $10 \times 2^9$.
Total pairs = $10 \times 512 = 5120$.
Ans. (5120)
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