Sets & RelationsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Sets & Relations: Let Number Elements Relation Divides (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

A=\{1,4,7\}

and

B=\{2,3,8\}

. The number of elements in the relation

R=\{((a_1,b_1),(a_2,b_2))\in(A\times B)\times(A\times B):a_1+b_2\text{ divides }a_2+b_1\}

Solution

Answer: 18

Step 1: List all achievable values of

a+b

for

(a,b)\in A\times B

and their frequencies The set

\{a+b : a\in A,\,b\in B\} = \{3,4,6,7,9,10,12,15\}

with frequencies:

c(3)=1,\ c(4)=1,\ c(6)=1,\ c(7)=1,\ c(9)=2,\ c(10)=1,\ c(12)=1,\ c(15)=1

. Step 2: For each divisor

u

, count valid multiples

v

and accumulate The total count equals

\displaystyle\sum_{\substack{u,v\,\in\,S \\ u\mid v}} c(u)\cdot c(v)

u=3

v\in\{3,6,9,12,15\}

\sum c(v)=1+1+2+1+1=6

. Contribution:

1\times6=6

u=4

v\in\{4,12\}

\sum c(v)=2

. Contribution:

1\times2=2

u=6

v\in\{6,12\}

\sum c(v)=2

. Contribution:

1\times2=2

u=7

v\in\{7\}

. Contribution:

1\times1=1

u=9

v\in\{9\}

. Contribution:

2\times2=4

u=10

v\in\{10\}

. Contribution:

1\times1=1

u=12

v\in\{12\}

. Contribution:

1\times1=1

u=15

v\in\{15\}

. Contribution:

1\times1=1

. Step 3: Sum all contributions

6+2+2+1+4+1+1+1 = 18

Answer: 18

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