JEE Main 2026 — Mathematics Algebra

Let $A = \{1, 4, 7\}$A={1,4,7} and $B = \{2, 3, 8\}$B={2,3,8}. The relation $R$R consists of pairs $((a_1, b_1), (a_2, b_2))$((a1​,b1​),(a2​,b2​)) from $A \times B$A×B such that $(a_1 + b_2)$(a1​+b2​) divides $(a_2 + b_1)$(a2​+b1​).

Let $x = a_1 + b_2$x=a1​+b2​ and $y = a_2 + b_1$y=a2​+b1​. Note that $(a_1, b_2)$(a1​,b2​) and $(a_2, b_1)$(a2​,b1​) are both independent elements of $A \times B$A×B. So choosing $((a_1, b_1), (a_2, b_2))$((a1​,b1​),(a2​,b2​)) is equivalent to independently choosing two elements of $A \times B$A×B with sums $x$x and $y$y such that $x \mid y$x∣y.

All possible sums $a + b$a+b with $a \in A$a∈A and $b \in B$b∈B:

$1 + 2 = 3$1+2=3, $1 + 3 = 4$1+3=4, $1 + 8 = 9$1+8=9

$4 + 2 = 6$4+2=6, $4 + 3 = 7$4+3=7, $4 + 8 = 12$4+8=12

$7 + 2 = 9$7+2=9, $7 + 3 = 10$7+3=10, $7 + 8 = 15$7+8=15

Frequencies of each sum:

$f(3) = 1$f(3)=1, $f(4) = 1$f(4)=1, $f(6) = 1$f(6)=1, $f(7) = 1$f(7)=1, $f(9) = 2$f(9)=2, $f(10) = 1$f(10)=1, $f(12) = 1$f(12)=1, $f(15) = 1$f(15)=1

For each possible divisor $x$x, count the number of valid pairs $(x, y)$(x,y) with $x \mid y$x∣y, each contributing $f(x) \cdot f(y)$f(x)⋅f(y):

$x = 3$x=3: divides $3, 6, 9, 12, 15$3,6,9,12,15

Ways $= 1 \cdot (1 + 1 + 2 + 1 + 1) = 6$=1⋅(1+1+2+1+1)=6

$x = 4$x=4: divides $4, 12$4,12

Ways $= 1 \cdot (1 + 1) = 2$=1⋅(1+1)=2

$x = 6$x=6: divides $6, 12$6,12

Ways $= 1 \cdot (1 + 1) = 2$=1⋅(1+1)=2

$x = 7$x=7: divides $7$7

Ways $= 1 \cdot 1 = 1$=1⋅1=1

$x = 9$x=9: divides $9$9

Ways $= 2 \cdot 2 = 4$=2⋅2=4

$x = 10$x=10: divides $10$10

Ways $= 1 \cdot 1 = 1$=1⋅1=1

$x = 12$x=12: divides $12$12

Ways $= 1 \cdot 1 = 1$=1⋅1=1

$x = 15$x=15: divides $15$15

Ways $= 1 \cdot 1 = 1$=1⋅1=1

Total number of elements in $R$R:

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

Hence, the answer is $18$18.