Solution:
If the number 12n, for any natural number n, ends with the digit 0 or 5, then it is divisible by 5. That is, the prime factorization of 12n contains the prime 5. This is not possible because prime factorisation of 12n = (22 x 3)n = 22n x 3n; so the only primes in the factorisation of 12n are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 12n. So, there is no natural number n for which 12n ends with the digit zero.
