JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q5

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MathematicsSingle correct+4 / -1Question 5 of 75

Let $A$ be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let $B$ be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in $A\cap B$ , which are divisible by 3, is :

Answer and worked solution

Official answer: B

Terms. $A=\{1,6,\dots,501\}$ (numbers $\equiv1\pmod5$ ) and $B=\{9,16,\dots,499\}$ (numbers $\equiv2\pmod7$ ).

Common terms. $x\equiv1\pmod5$ and $x\equiv2\pmod7$ give $x\equiv16\pmod{35}$ : $x=16+35t\le499$ , so $t=0,\dots,13$ .

Divisible by 3. $16+35t\equiv1+2t\equiv0\pmod3$ , so $t\equiv1\pmod3$ : $t=1,4,7,10,13$ . That is 5 numbers.

(B)

Topics:Arithmetic progression Sets and operations

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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 5