Show that 6n can not end with digit 0 for any natural number 'n'.

If $6^n$ ends with digit $0$, it would be divisible by $5$. So, prime factorization of $6^n$ would contain $5$. But $6^n = (2 \times 3)^n$, the only prime factorization of $6^n$ are $2$ and $3$ as per fundamental theorem of Arithmetic . There is no other prime in the factorization of $6^n$. So, there is no natural number $n$ for which $6^n$ ends with digit zero.