Question

If \(2.4^{2n+1} + 3^{3n+1}\) is divisible by \(k\) for all \(n \in \mathbb{N}\), then \(k\) is:

• A. \(209\)
• B. \(11\)
• C. \(8\)
• D. \(3\)

Hint:

For expressions involving powers and divisibility, it's effective to test divisibility using small values of \(n\) and calculate possible divisors from the given options. This helps detect patterns that may hold for all \(n\).

Answer 1

The Correct Option is B

Step 1: Consider the expression \(2^{4n+1} + 3^{3n+1}\). 
- We observe that the powers of 2 and 3 are increasing with \(n\). The expression is defined for all \(n \in \mathbb{N}\), so we will check the divisibility for specific values of \(n\) to detect a pattern. 
Step 2: Check for divisibility by smaller numbers for base cases (e.g., \(n = 1\)). 
- For \(n = 1\), we compute: \[ 2^{4\cdot1+1} + 3^{3\cdot1+1} = 2^5 + 3^4 = 32 + 81 = 113. \] Now, we will check the divisibility of \(113\) by the options given. 
Step 3: Test \(113\) for divisibility by the options given. 
- \(113 \mod 209 = 113\) (not divisible)
- \(113 \mod 11 = 3\) (divisible)
- \(113 \mod 8 = 1\) (not divisible)
- \(113 \mod 3 = 2\) (not divisible) 
Step 4: Since \(113\) is divisible by \(11\) and considering the powers involved, \(11\) is a likely candidate for \(k\). Therefore, the value of \(k\) is \(11\).