Mathematics · Sets and Relations

JEE Main 2026 — 5 April, Evening Shift — Question 44

Let $A = {1,4,7}$ and $B = {2,3,8}.$ Then the number of elements in the relation $R = ((a₁,b₁),(a₂,b₂)) ∈ (A×B)×(A×B) : a_1 + b_2$ divides $a_2 + b_1$ is ______.

Answer: 18

Numerical answer — enter this value.

Step-by-step solution

Let $A = \{1,4,7\}$ , $B = \{2,3,8\}$ . Then $A \times B$ has $3 \times 3 = 9$ ordered pairs.

We need ordered pairs $((a_1,b_1),(a_2,b_2))$ such that $a_1 + b_2$ divides $a_2 + b_1$ .

List all pairs $(a,b)$ : $(1,2), (1,3), (1,8), (4,2), (4,3), (4,8), (7,2), (7,3), (7,8)$ .

For each $(a_1,b_1)$ , compute $a_2+b_1$ for all $(a_2,b_2)$ and check divisibility by $a_1+b_2$ .

Systematic counting yields 18 ordered pairs satisfying the condition.

Thus, the number of elements in $R$ is $\boxed{18}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Sets and Relations
Topic: Relations