Answer :

If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorization must include primes 2 and 5 both as 10 = 2 ×5

Prime factorization of 6^{n} = (2 ×3) ^{n}

In the above equation it is observed that 5 is not in the prime factorization of 6^{n}

By Fundamental Theorem of Arithmetic Prime factorization of a number is unique. So 5 is not a prime factor of 6^{n}.

Hence, for any value of n, 6^{n} will not be divisible by 5.

Therefore, 6^{n} cannot end with the digit 0 for any natural number n.

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