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

If the number $6^n$, for any natural number n, ends with digit 0, then it would be divisible by 5. That is, the prime factorization of $6^n$ would contain the prime number 5. This is not possible because $6^n=(2\times 3{)}^n=2^n\times 3^n$ ; so the only primes in the factorization of $6^n$ are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of $6^n$. So, there is no natural number n for which $6^n$ ends with the digit zero.