Show that $12^{n}$ cannot end with the digit $0$ or $5$ for any natural number $n$.

(N/A) If any number ends with the digit $0$ or $5$,it is always divisible by $5$.
If $12^{n}$ ends with the digit $0$ or $5$,it must be divisible by $5$.
This is possible only if the prime factorization of $12^{n}$ contains the prime number $5$.
Now,the prime factorization of $12$ is $12 = 2^{2} \times 3$.
Therefore,$12^{n} = (2^{2} \times 3)^{n} = 2^{2n} \times 3^{n}$.
By the Fundamental Theorem of Arithmetic,the prime factorization of $12^{n}$ is unique and contains only the prime factors $2$ and $3$.
Since $5$ is not a prime factor of $12^{n}$,$12^{n}$ cannot be divisible by $5$ for any natural number $n$.
Hence,$12^{n}$ cannot end with the digit $0$ or $5$ for any natural number $n$.