JEE Mathematics 2025 Question with Solution

Given: $S = \{p_1, p_2, \dots, p_{10}\}$ is the set of the first ten prime numbers. $P$ is the set of all possible products of distinct elements of $S$, and $A = S \cup P$.

Find: The number of ordered pairs $(x,y)$ such that $x \in S$, $y \in A$, and $x$ divides $y$.

Each element of $A$ corresponds to a non-empty subset of $S$:

• a one-element subset gives an element of $S$,
• a larger subset gives a product in $P$.

So the total number of elements in $A$ is

$$|A| = 2^{10} - 1 = 1023$$

because all non-empty subsets of a set of $10$ elements are counted.

Now fix any $x \in S$. An element $y \in A$ is divisible by $x$ exactly when the corresponding subset contains $x$. After fixing $x$, each of the remaining $9$ primes may be either chosen or not chosen independently.

Hence the number of elements of $A$ divisible by a fixed $x$ is

$$2^9 = 512$$

Since there are $10$ choices for $x \in S$, the total number of ordered pairs is

$$10 \times 512 = 5120$$

Therefore, the number of ordered pairs is $5120$.

Given: $S$ has $10$ distinct prime numbers.

Find: Count all ordered pairs $(x,y)$ with $x \in S$, $y \in A$, and $x \mid y$.

Write every element of $A$ as the product of elements from a non-empty subset of $S$. This works because:

• if the subset has size $1$, the product is a prime in $S$,
• if the subset has size at least $2$, the product belongs to $P$.

So every non-empty subset of $S$ gives exactly one element of $A$.

For a fixed prime $x \in S$, we want those subsets whose product is divisible by $x$. That happens exactly when $x$ is included in the subset.

Once $x$ is included, the remaining $9$ primes can be chosen freely. Therefore the number of such subsets is

$$\sum_{k=0}^{9} \binom{9}{k} = 2^9 = 512$$

Thus, for each fixed $x$, there are $512$ possible values of $y$.

Finally,

$$\text{Total ordered pairs} = 10 \cdot 512 = 5120$$

So the required answer is $5120$.

Note: The alternate working shown on the page writes $1 + 2^9 = 513$ for each fixed prime, but then concludes $5120$. The correct count is $2^9 = 512$ because the one-element subset corresponding to $x$ itself is already included in these $2^9$ cases.