Maths Q7 – JEE-MAIN 2026

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 :

The number of triangles formed by joining the vertices of an n-sided polygon is equivalent to choosing 3 vertices out of n.

Step 1: Express

p_{n}

and use the given relation✦ Active

The total number of triangles formed by joining the vertices of an n-side regular polygon is given by the combination formula:

p_{n} = (\binom{n}{3}) = \frac{n (n - 1) (n - 2)}{3 \cdot 2 \cdot 1}

We are given the relation $p_{n + 1} - p_{n} = 66$ . Substituting the formula for $p_{n}$ :

(\binom{n + 1}{3}) - (\binom{n}{3}) = 66

Using Pascal's identity, $(\binom{k}{r}) - (\binom{k - 1}{r}) = (\binom{k - 1}{r - 1})$ , with $k = n + 1$ and $r = 3$ , we get:

(\binom{n + 1}{3}) - (\binom{n}{3}) = (\binom{n}{2})

So, the equation becomes:

(\binom{n}{2}) = 66

Step 2: Solve for n○ Expand

Expand the combination formula for $(\binom{n}{2})$ :

\frac{n (n - 1)}{2} = 66

Multiply both sides by 2:

n (n - 1) = 132

We need to find two consecutive integers whose product is 132. By inspection, $12 \times 11 = 132$ . Therefore, $n = 12$ .

💡 Teacher's Secret Hint

Remember that $n$ must be an integer greater than or equal to 3 for $(\binom{n}{3})$ to be defined.

Step 3: Find the sum of distinct prime divisors of n○ Expand

The value of $n$ is 12. Now, we find the prime factorization of 12:

12 = 2^{2} \times 3

The distinct prime divisors of 12 are 2 and 3. The sum of these distinct prime divisors is:

2 + 3 = 5

💡 Teacher's Secret Hint

Ensure you only sum the *distinct* prime divisors, not all prime factors with their multiplicities.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 7 – JEE-MAIN 2026