For what at least value of n where n is a natural number is: $19^{n}$ is divisible by 9.

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Solution

Definition:
A prime number is a number that has only two factors and is divisible by itself and 1.
Factorization:
$19 = 1 \times 19$
As $19$ is a prime number,
$19$ is not divisible by $9$ .
Hence, $9$ and $19$ are co-primes.
Therefore, for no value of n, $19^{n}$ is divisible by 9.

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