Question

If \( 3^{2n+2} - 8n - 9 \) is divisible by \( 2^p \) \(\forall n \in \mathbb{N}\), then the maximum value of \( p \) is

• A. \( 8 \)
• B. \( 7 \)
• C. \( 6 \)
• D. \( 9 \)
  \

Hint:

For divisibility problems involving exponents, analyze the cyclic nature of modular powers and use patterns to determine higher power divisibility.

Answer 1

The Correct Option is C

We need to determine the maximum value of \( p \) such that \( 3^{2n+2} - 8n - 9 \) is divisible by \( 2^p \) for all \( n \in \mathbb{N} \). Step 1: Consider divisibility by \( 2^p \)
For divisibility by \( 2^p \), we analyze \( 3^{2n+2} \mod 2^p \). Using the pattern: \[ 3^2 = 9 \equiv 1 \pmod{8} \] Thus, for all \( n \geq 1 \), \[ 3^{2n+2} = (3^2)^{n+1} \equiv 1^{n+1} \equiv 1 \pmod{8}. \] Step 2: Evaluating \( 3^{2n+2} - 8n - 9 \) modulo \( 2^p \)
\[ 3^{2n+2} - 8n - 9 \equiv 1 - 8n - 9 \pmod{8} \] \[ \equiv -8n - 8 \equiv -8(n+1) \pmod{8}. \] Since \( -8(n+1) \) is always divisible by 8, we extend our analysis for higher powers of 2. Step 3: Finding the maximum \( p \)
By further checking divisibility by \( 2^p \), we find that the maximum possible value of \( p \) satisfying the given condition is: \[ p = 6. \] \bigskip