If $n$ is a natural number, then $9^{2n} - 4^{2n}$ is always divisible by:
$5$ only
$13$ only
Both $5$ and $13$
Neither $5$ nor $13$
Step-by-Step Solution
Key Concept: $a^n - b^n$ is divisible by $(a - b)$ for all $n$, and by $(a + b)$ for even $n$.
Stepwise Solution:
Here $9^{2n} - 4^{2n} = (9^2)^n - (4^2)^n = 81^n - 16^n$. [0.5 Mark]
$81^n - 16^n$ is divisible by $81 - 16 = 65$. Since $65 = 5 \times 13$, it is divisible by both $5$ and $13$. [0.5 Mark]
Marking Scheme:
• Expressing as $81^n - 16^n$: 0.5 Mark
• Divisibility by 65 ($5 \times 13$): 0.5 Mark
Correct Answer: Both $5$ and $13$