Question

A regular polygon with \( n \) sides is given. \( P_n \) denotes number of triangles formed by joining any three points of given regular polygon. If \( P_{n+1} - P_n = 66 \), then the sum of all prime divisors of \( n \) is

• A. 9
• B. 5
• C. 11
• D. 23

Hint:

For polygon vertex problems, the number of triangles formed by choosing any three vertices is always \( \binom{n}{3} \). Also remember the useful identity \( \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2} \).

Answer 1

The Correct Option is B

Step 1: Write the formula for \( P_n \).
The number of triangles formed by joining any three vertices of an \( n \)-sided polygon is
\[ P_n = \binom{n}{3} = \frac{n(n-1)(n-2)}{6}. \] Similarly,
\[ P_{n+1} = \binom{n+1}{3} = \frac{(n+1)n(n-1)}{6}. \]
Step 2: Find \( P_{n+1} - P_n \).
Now,
\[ P_{n+1} - P_n = \binom{n+1}{3} - \binom{n}{3}. \] Using the identity
\[ \binom{n+1}{3} - \binom{n}{3} = \binom{n}{2}, \] we get
\[ P_{n+1} - P_n = \frac{n(n-1)}{2}. \] Given that
\[ P_{n+1} - P_n = 66, \] so
\[ \frac{n(n-1)}{2} = 66. \]
Step 3: Solve for \( n \).
Multiplying both sides by \( 2 \), we get
\[ n(n-1) = 132. \] So,
\[ n^2 - n - 132 = 0. \] Factorizing,
\[ n^2 - 12n + 11n - 132 = 0 \] \[ n(n-12) + 11(n-12) = 0 \] \[ (n+11)(n-12)=0. \] Since \( n \) is positive,
\[ n=12. \]
Step 4: Find the sum of all prime divisors of \( n \).
Prime divisors of \( 12 \) are \( 2 \) and \( 3 \).
Therefore, their sum is
\[ 2+3=5. \]
Final Answer: \( 5 \)