${{10}^{n}}+{{3}^{n}}+2=1\underbrace{00...00}_{n-1}2+{{3}^{n}}$ we must have 2+ the last digit of the ${{3}^{n}}$ together 1. So n=4k+2, but must have k=0, because else we have too much zero after 1, and no zeros at the end. So only situation it has. Now you've find that!