JEE Main 2026 — Mathematics Algebra

The set $A$A consists of the first 101 terms of an A.P. with first term $a_1 = 1$a1​=1 and common difference $d_1 = 5$d1​=5.

The terms of $A$A are $1, 6, 11, 16, \dots, 1 + (101-1) \times 5 = 501$1,6,11,16,…,1+(101−1)×5=501.

The set $B$B consists of the first 71 terms of an A.P. with first term $b_1 = 9$b1​=9 and common difference $d_2 = 7$d2​=7.

The terms of $B$B are $9, 16, 23, 30, \dots, 9 + (71-1) \times 7 = 499$9,16,23,30,…,9+(71−1)×7=499.

The first common term of $A$A and $B$B is $16$16.

The common difference of the terms in $A \cap B$A∩B is the least common multiple of $d_1$d1​ and $d_2$d2​, which is $\text{lcm}(5, 7) = 35$lcm(5,7)=35.

The general term of $A \cap B$A∩B is given by $T_k = 16 + 35k$Tk​=16+35k, where $k \ge 0$k≥0 is an integer.

Since the maximum possible value in $A \cap B$A∩B cannot exceed the maximum value in $B$B (which is $499$499), we have:

$$16 + 35k \le 499$$16+35k≤499

$$35k \le 483$$35k≤483

$$k \le 13.8$$k≤13.8

Thus, the possible values for $k$k are $0, 1, 2, \dots, 13$0,1,2,…,13.

We require the terms in $A \cap B$A∩B to be divisible by $3$3. Therefore:

$$16 + 35k \equiv 0 \pmod 3$$16+35k≡0(mod3)

$$1 + 2k \equiv 0 \pmod 3$$1+2k≡0(mod3)

$$2k \equiv 2 \pmod 3$$2k≡2(mod3)

$$k \equiv 1 \pmod 3$$k≡1(mod3)

The values of $k$k in the range $0 \le k \le 13$0≤k≤13 that satisfy $k \equiv 1 \pmod 3$k≡1(mod3) are $1, 4, 7, 10, 13$1,4,7,10,13.

There are exactly $5$5 such values of $k$k, which correspond to $5$5 elements in $A \cap B$A∩B that are divisible by $3$3.

Answer: $5$5