Let $abc$ be the three-digit base $11$ number, where $a,$ $b,$ and $c$ are the digits. This number is equal to $121a + 11b + c.$
Then the number in base $6$ is $100a + 10b + c,$ since the digits are in reverse order.
We have the following equations:
$$121a + 11b + c = 100a + 10b + c.$$
Simplifying, we get $21a = b.$
Also, $a < 6,$ since $N$ is a three-digit base $11$ number. Furthermore, $b < 6$ and $c < 6.$
But since $b = 21a,$ we have $b = 0,$ $b = 21,$ $b = 42.$
Since $b < 6,$ the only possible value is $b = 42,$ which means $a = 2.$ Therefore, $c = 1,$ and $N = 211_{11} = \boxed{257}$ in decimal.
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 $6$, it has the same three digits, in reverse order. What is $N$? (Express your answer in decimal.)
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