Mathematics · Permutations and Combinations

JEE Main 2026 — 2 April, Evening Shift — Question 29

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

Option A:
7
Option B:
8
Option C:
5
Correct
Option D:
6

Answer: C

Step-by-step solution

$\mathrm{P}_{\mathrm{n}}={ }^{\mathrm{n}} \mathrm{C}_{3}$ $P_{n+1}-P_{n}=66$ ${ }^{\mathrm{n}+1} \mathrm{C}_{3}-{ }^{\mathrm{n}} \mathrm{C}_{3}=66$ $\Rightarrow \frac{(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)}{6}-\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{6}=66$ $\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{6}[\mathrm{n}+1-\mathrm{n}+2]=66$ $\Rightarrow \mathrm{n}(\mathrm{n}-1)=132$ $\mathrm{n}=12$ Prime divisors of $12$ are $2$ & $3$ sum $=2+3=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Permutations and Combinations
Topic: Combinations