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.