Question

If $n$is a natural number then $9^{2n}-4^{2n}$is always divisible by

• A. $5$
• B. $13$
• C. both $5$and $13$
• D. None of these

Answer 1

The correct option is C

both $5$and $13$

Solve for divisor of given expression

Given, $9^{2n}-4^{2n}$is in the form of $a^{2n}-b^{2n}$ where $n$ is a natural number.

Since $a^{2n}-b^{2n}$is always divisible by $(a-b)$ and $(a+b)$,

$9^{2n}-4^{2n}$is always divisible by:

$$\begin{gathered} 9-4=5 \\ 9+4=13 \end{gathered}$$

$\therefore$$9^{2n}-4^{2n}$is always divisible by both $5$ and$13$

Hence the correct option is option(C) i.e. both $5$ and$13$