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Common AP Terms Divisible by 3: Count = 5 | JEE Main 2026

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

Question
Let AA be the set of first 101101 terms of an A.P. whose first term is 11 and common difference is 55, and let BB be the set of first 7171 terms of an A.P. whose first term is 99 and common difference is 77. Then the number of elements in A∩BA\cap B which are divisible by 33 is
A44
B55correct
C66
D77
Solution
Step 1: Write the two sets:
A={1,6,11,16,…} (101 terms, last term 1+100⋅5=501),A=\{1,6,11,16,\ldots\}\ \text{(101 terms, last term } 1+100\cdot5=501),
B={9,16,23,…} (71 terms, last term 9+70⋅7=499).B=\{9,16,23,\ldots\}\ \text{(71 terms, last term } 9+70\cdot7=499).
 Step 2: Common terms form an A.P. with common difference lcm(5,7)=35\mathrm{lcm}(5,7)=35. The first common term is 1616. Step 3: Common terms: 16+35k≤49916+35k\le 499 (and ≤501\le 501), i.e. k≤13.8k\le 13.8, so k=0,…,13k=0,\ldots,13 — 14 terms:
A∩B={16,51,86,121,156,191,226,261,296,331,366,401,436,471}.A\cap B=\{16,51,86,121,156,191,226,261,296,331,366,401,436,471\}.
 Step 4: Among these, the ones divisible by 33:
{51,156,261,366,471} ⇒ 5 terms.\{51,156,261,366,471\}\ \Rightarrow\ 5\ \text{terms}.
 Correct answer: (2)
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