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

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Step 1: If $6^n$ ends with digit $0$, it would be divisible by $5$. So, prime factorization of $6^n$ would contain $5$. But $6^n = (2 	imes 3)^n$, the only prime factors of $6^n$ are $2$ and $3$ as per fundamental theorem of Arithmetic.
Step 2: 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.

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