Let $N$ be a positive integer. The number $N$ has three digits when expressed in base $11.$ When the number $N$ is expressed in base $5$, it has the same three digits, in reverse order. What is $N$? (Express your answer in decimal.)

Answer 1

Let $N = \overline{abc}_{11},$ where $a,$ $b,$ and $c$ are the digits of $N$ in base $11.$ Then
\begin{align*}
N &= a \cdot 11^2 + b \cdot 11 + c \\
&= (a \cdot 5^2 + b \cdot 5 + c)_{11} \\
&= 25a + 5b + c.
\end{align*}Then $25a + 5b + c = \overline{cba}_5.$

From the conditions of the problem, $N$ is a three-digit number in base $11,$ so $a$ is nonzero. Thus, $a,$ $b,$ and $c$ are digits that satisfy the equations
\begin{align*}
25a + 5b + c &= 5^2 c + 5b + a \\
5^2 c + 5b + a &\le 10^2 + 10 + 1,
\end{align*}which simplify to $20a = 24c$ and $100c + 5b + a \le 111.$ Since $c$ divides 20, the only possible value for $c$ is 4. Then $a = 5,$ and $c = 4,$ which means $b = 2.$ Thus, $N = \boxed{542}.$