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Answer :

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.

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