Answer :

If any number end with digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as

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

It can be observed that 5 is not in the prime factorization of 6^{n}.

Hence, for an value of n, 6^{n} will not visible by 5.

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

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