JEE Main 2026 Apr 2 Shift 2, Mathematics Q7: Combinations

JEE Main2026Apr 2, Shift 2Mathematics

Let $p_{n}$ denote the total number of triangles formed by joining the vertices of an $n$ -side regular polygon. If $p_{n + 1} - p_{n} = 66$ , then the sum of all distinct prime divisors of $n$ is :

A
$7$
B
$8$
C
$5$
D
$6$

Solution

The number of triangles is $p_{n} =^{n} C_{3}$ . The identity $^{n + 1} C_{3} -^{n} C_{3} =^{n} C_{2}$ gives

$^{n} C_{2} = \frac{n ( n - 1 )}{2} = 66$ , so $n (n - 1) = 132$ , yielding $n = 12$ .

The prime factorisation $12 = 2^{2} \cdot 3$ has distinct prime divisors $2$ and $3$ , whose sum is $5$ .

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JEE Main 2026 Apr 2 Shift 2, Mathematics Q7: Combinations

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