If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorization must include primes 2 and 5 both as 10 = 2 ×5
Prime factorization of 6n = (2 ×3) n
In the above equation it is observed that 5 is not in the prime factorization of 6n
By Fundamental Theorem of Arithmetic Prime factorization of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Rate this question :