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JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2026 · 2 April · Evening Shift · Mathematics · Permutation and Combination

Let p_n denote the total number of triangles formed by joining the...

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

Options

  1. A77
  2. B88
  3. C55
  4. D66

Correct answer

C. 55

Step-by-step solution

The number of triangles formed by joining the vertices of an nn-sided regular polygon is given by pn=nC3p_n = ^{n}C_{3}.

Given that pn+1−pn=66p_{n+1} - p_n = 66.

n+1C3−nC3=66^{n+1}C_{3} - ^{n}C_{3} = 66

Using the property n+1Cr−nCr=nCr−1^{n+1}C_{r} - ^{n}C_{r} = ^{n}C_{r-1}, we get:

nC2=66^{n}C_{2} = 66

n(n−1)2=66\dfrac{n(n-1)}{2} = 66

n(n−1)=132n(n-1) = 132

n2−n−132=0n^2 - n - 132 = 0

(n−12)(n+11)=0(n-12)(n+11) = 0

Since nn must be a positive integer, n=12n = 12.

The prime factorization of 1212 is 22×32^2 \times 3. The distinct prime divisors of 1212 are 22 and 33.

The sum of all distinct prime divisors of nn is 2+3=52 + 3 = 5.

Answer: 55
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