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

A number ending with 0 is divisible by 10 according to the divisibility rule

And a number divisible by 10 is also divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of $$6^n = (2 × 3)^n$$

5 is not in the prime factorisation of $$6^n$$
So, for any value of n, $$6^n$$ will not be divisible by 5

Hence, $$6^n$$ cannot end with the digit 0 for any natural number n.