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 6n = (2×3)n
It can be observed that 5 is not in the prime factorization of 6n.
Hence, for an value of n, 6n will not visible by 5.
∴ 6n cannot end with the digit 0 for any natural number n.
Rate this question :