${{10}^{n}}+{{3}^{n}}+2=1\underbrace{00...00}_{n-1}2+{{3}^{n}}$ 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 siruation it is n=2, the palindrom number it is 111

Mr.Math.Boy wrote: because else we have too much zero after 1 True but that has to be proven. Otherwise the solution is not complete.

The numbers of 0 on the LHS i creasing fast with every n power, much that the 0 in the pover of 3,where we have not consecutive 0 etc.

Mr.Math.Boy wrote: The numbers of 0 on the LHS i creasing fast with every n power, much that the 0 in the pover of 3,where we have not consecutive 0 etc. You have to prove that mathematically.

First, $n = 2$ works and $n = 1$ does not. For $n\ge 3$, we must have $n - \lfloor \log_{10}\left( 3^n\right)\rfloor - 1 \le \lfloor \log_{10}\left( 3^n\right)\rfloor - 1$ where the LHS is the number of zeroes before the first digit of $3^n+2$ and the RHS is an upper bound on the number of zeroes in the number $3^n + 2$. However, this is false because $n > 2n\log_{10}\left( 3\right) = 2\log_{10}\left( 3^n\right) \ge 2\lfloor \log_{10}\left( 3^n\right) \rfloor $

The number begins with a $1$, so it must end with a $1$. $3^n+2 \equiv 1 \pmod{10}$ $3^n \equiv 9 \pmod{10}$ $n \equiv 2 \pmod{4}$ for $n=2$, $10^2+3^2+2=111$, which is a palindrome. However, for $n>2$, we would also need $3^n+2 \equiv 1 \pmod{100}$ to account for the $0$s contributed by $10^n$ $3^n \equiv 99 \pmod{100}$ This implies that $3^n \equiv 99 \pmod{4}$. $3^n \equiv 3 \pmod{4}$ implies $n$ is odd But $n \equiv 2 \pmod{4}$, so it is even, contradiction. Only $\boxed{2}$ works.