Question:
If $n$ is a natural number, then $9^{2 n}-4^{2 n}$ is always divisible by
(a) 5
(b) 13
(c) both 5 and 13
(d) None of these
[Hint : $9^{2 n}-4^{2 n}$ is of the form $a^{2 n}-b^{2 n}$ which is divisible by both $a-b$ and $a+b$. So, $9^{2 n}-4^{2 n}$ is divisible by both $9-$ $4=5$ and $9+4=13$.]
Solution:
We know that $a^{2 n}-b^{2 n}$ is always divisible by both $a-b$ and $a+b$.
So, $9^{2 n}-4^{2 n}$ is always divisible by both $9-4=5$ and $9+5=13$.
Hence, the correct choice is $(c)$.
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