Let p_n denote the total number of triangles formed by joining the vertices of an n-side r…

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 :

Accepted Answer

${P}_{n}={}^{n}{C}_{3}$

In an $n$-sided polygon, any $3$ vertices form a triangle, so the total number of triangles is ${P}_{n}={}^{n}{C}_{3}$.

${P}_{n+1}-{P}_{n}=66$

This means when we increase the number of sides from $n$ to $n+1$, the number of triangles increases by $66$.

${}^{n+1}{C}_{3}-{}^{n}{C}_{3}=66$

Now we write both combinations using the formula ${}^{n}{C}_{3}=\frac{n(n-1)(n-2)}{6}$.

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

Take the common factor $\frac{n(n-1)}{6}$ from both terms.

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

Inside the bracket, simplify: $n+1-n+2=3$.

$\Rightarrow n(n-1)=132$

So we need two consecutive integers whose product is $132$. That gives $n=12$ because $12\cdot 11=132$.

$n=12$

Prime divisors of 12 are 2 \& 3

The distinct prime divisors of $12$ are $2$ and $3$.

Sum $=2+3=5$

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