Let denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If , then the sum of all distinct prime divisors of n is: [2026]

Question

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: [2026]

Smart Achievers · Accepted Answer

(3)

$p_{n} = {}^{n}C_{3}$

We have, $p_{n + 1} - p_{n} = 66$

${}^{n + 1}C_{3} - {}^{n}C_{3} = 66$

$\Rightarrow \frac{(n + 1) (n) (n - 1)}{6} - \frac{n (n - 1) (n - 2)}{6} = 66$

$\Rightarrow n (n - 1) [(n + 1) - (n - 2)] = 396$

$\Rightarrow 3 n (n - 1) = 396 \Rightarrow n^{2} - n - 132 = 0$

$\Rightarrow n (n - 12) (n + 11) = 0 \Rightarrow n = 12, - 11$

We know, $12 = 2^{2} 3^{1}$

$∴$ Divisor are 1, 2, 3, 4, 6, 12

Prime divisor are 2 and 3

$∴$ Sum = 5

Solution

Q.
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: [2026]

1 7

2 8

3 5

4 6

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Solution

Q. Let pn denote the total number of triangles formed by joining the vertices of an n-side regular polygon. If pn+1–pn=66, then the sum of all distinct prime divisors of n is: [2026]

1 7

2 8

3 5

4 6

Ans. (3) pn=C3n We have, pn+1–pn=66 C3n+1–C3n=66 ⇒ (n+1)(n)(n–1)6–n(n–1)(n–2)6=66 ⇒ n(n–1)[(n+1)–(n–2)]=396 ⇒ 3n(n–1)=396 ⇒ n2–n–132=0 ⇒ n(n–12)(n+11)=0 ⇒ n=12, –11 We know, 12=2231 ∴ Divisor are 1, 2, 3, 4, 6, 12 Prime divisor are 2 and 3 ∴ Sum = 5

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Q.
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: [2026]