Check whether $6^{n}$ can end with the digit $0$ for any natural number $n$.

(N/A) If any number ends with the digit $0,$ it must be divisible by $10.$ This implies that it must also be divisible by both $2$ and $5,$ since $10 = 2 \times 5.$
The prime factorisation of $6^{n}$ is $(2 \times 3)^{n} = 2^{n} \times 3^{n}.$
It can be observed that the prime factor $5$ is not present in the prime factorisation of $6^{n}.$
Since $5$ is not a factor of $6^{n},$ $6^{n}$ is not divisible by $5$ for any natural number $n.$
Therefore,$6^{n}$ cannot end with the digit $0$ for any natural number $n.$