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

Steps to Solve

1. Prime Factorization of 6 Find the prime factors of 6: $6 = 2 \times 3$.

2. Expressing $6^n$ in terms of its prime factors Express $6^n$ using its prime factors: $6^n = (2 \times 3)^n = 2^n \times 3^n$.

3. Condition for a number to end with 0 A number ends with the digit 0 if it is divisible by 10. The prime factorization of 10 is $10 = 2 \times 5$. Thus, for a number to end in 0, its prime factorization must contain both 2 and 5.

4. Checking for the presence of 5 in the prime factorization of $6^n$ The prime factorization of $6^n$ is $2^n \times 3^n$. It contains the prime factor 2 but does not contain the prime factor 5 for any natural number $n$.

5. Conclusion Since the prime factorization of $6^n$ does not include the prime factor 5, $6^n$ cannot be divisible by 10, and therefore, $6^n$ cannot end with the digit 0 for any natural number $n$.