Mathematics · Sets and Relations

JEE Main 2025 — 24 January, Morning Shift — Question 25

Let S={p1,p2……,p10}S=\left\{p_{1}, p_{2} \ldots \ldots, p_{10}\right\} be the set of first ten prime numbers. Let A=S∪PA=S \cup P, where PP is the set

of all possible products of distinct element of SS. Then the number of all ordered pairs ( x,yx, y ), x∈Sx \in S, y∈Ay \in A,

such that xx divides yy, is _____\_\_\_\_\_ .

Answer: 5120

Numerical answer — enter this value.

Step-by-step solution

Let yx=λ\frac{y}{x}=\lambda

y=λxy=\lambda x =10×(9C0+9C1+9C2+9C3+….+9C9)=10 \times\left({ }^{9} \mathrm{C}_{0}+{ }^{9} \mathrm{C}_{1}+{ }^{9} \mathrm{C}_{2}+{ }^{9} \mathrm{C}_{3}+\ldots .+{ }^{9} \mathrm{C}_{9}\right)

=10×(29)=10 \times\left(2^{9}\right)

10×51210 \times 512

5120

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Sets and Relations
Topic
Sets