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:

• A. 7
• B. 8
• C. 5
• D. 6

Answer 1

The Correct Option is D

Step 1: Understanding the formula for \( p_n \).
The total number of triangles formed by joining the vertices of an \( n \)-sided polygon is given by the combination formula \( p_n = \binom{n}{3} \), as we need to select 3 vertices out of \( n \) vertices to form a triangle. This can be expressed as: \[ p_n = \frac{n(n-1)(n-2)}{6} \]
Step 2: Using the given condition.
We are given that \( p_{n+1} - p_n = 66 \), so we calculate: \[ p_{n+1} = \frac{(n+1)n(n-1)}{6} \] Thus: \[ p_{n+1} - p_n = \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} \] Simplifying: \[ p_{n+1} - p_n = \frac{n(n-1)[(n+1) - (n-2)]}{6} = \frac{n(n-1)(3)}{6} = \frac{n(n-1)}{2} \] We are given that \( p_{n+1} - p_n = 66 \), so: \[ \frac{n(n-1)}{2} = 66 \] Multiplying both sides by 2: \[ n(n-1) = 132 \] Solving for \( n \), we find that \( n = 12 \).
Step 3: Finding the sum of distinct prime divisors of \( n \).
The prime factorization of \( n = 12 \) is: \[ 12 = 2^2 \times 3 \] The distinct prime divisors of 12 are 2 and 3. The sum of these prime divisors is: \[ 2 + 3 = 5 \]
Step 4: Conclusion.
The sum of all distinct prime divisors of \( n \) is \( 2 + 3 = 5 \).
Final Answer: (D) 6