JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Permutations and Combinations

$S=\{p_1,\dots ,p_{10}\}$ consists of the first ten primes, and $A=S\cup P$ where $P$ is the set of all products of distinct elements of $S$; since a single prime is itself such a "product," $A$ is exactly the set of products of nonempty subsets of $S$ — one value for each of the $2^{10}-1$ nonempty subsets, all distinct because the elements are distinct primes.

Fix $x\in S$. An element $y\in A$ is divisible by $x$ exactly when the subset producing $y$ contains $x$. The number of subsets of $S$ containing a fixed element equals the number of subsets of the remaining $9$ primes, which is $2^9=512$ (including the subset $\{x\}$ itself, giving $y=x$).

So for each choice of $x$ there are $512$ valid $y$. With $10$ choices for $x$,