256. Palindromic Squares. A certain square is written in the base-6 numeral system using 5 different non-zero digits. If the digit in the units place is moved from the end of this number to the beginning, the square root of the resulting number matches the square root of the original number, written in reverse order. Find the original number.
256. $12345 \leqslant N^{2} \leqslant 54321$, so $113 \leqslant N \leqslant 221$. If we write the digits of the number $N$ in reverse order, the resulting number will also satisfy these inequalities. The number $N^{2}$ is divisible by 5, so $N$ is also divisible by 5. Let's list all three-digit numbers divisible by 5 and satisfying these inequalities: $113, 122, 131, 140, 145, 154, 203, 212$ and 221. Among them, there is only one pair of palindromes. By squaring it, we confirm that it provides the only solution to our problem: $(221)^{2}=53241, 15324=(122)^{2}$.

As a "free bonus," we find that $(203)^{2}=42013$ - another permutation of five consecutive digits.