JEE Main 2025 Jan 24 Shift 1, Mathematics Q25: Set

JEE Main2025Jan 24, Shift 1Mathematics

Let $S = {p_{1}, p_{2}, \dots, p_{10}}$ 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 ______.

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Solution

Fix $x \in S$ (10 choices). Count $y \in A$ such that $x ∣ y$ .

Every element of $A$ is a product of a non-empty subset of $S$ (single primes $\Rightarrow$ in $S$ ; products of 2 or more $\Rightarrow$ in $P$ ). So $A$ is in bijection with the non-empty subsets of $S$ .

An element $y = \prod_{i \in T} p_{i}$ (for some non-empty $T \subseteq S$ ) is divisible by $x = p_{k}$ iff $p_{k} \in T$ . The remaining 9 primes may each be present or absent freely, giving $2^{9} = 512$ subsets $T$ containing $p_{k}$ (none is empty since $p_{k}$ is already in).

So for each $x$ , there are 512 valid $y$ . Total ordered pairs $= 10 \times 512 = 5120$ .

$∴$ Answer: $5120$ .

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JEE Main 2025 Jan 24 Shift 1, Mathematics Q25: Set

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