**Solution: **

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