Question

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

• A. $13$
• B. Both $5$ and $13$
• C. $5$
• D. None of the above

Answer 1

The correct option is B Both $5$ and $13$

${81}^n-{16}^n\Rightarrow a^n-b^n$
is always divisible by $(a-b)$
$(81-16)=65$
$\therefore$ So factors are both $5$ and $13$