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 $12$, it has the same three digits, in reverse order. What is $N$? (Express your answer in decimal.)

Answer 1

Let $N$ have three digits $abc$ in base $11$, where $a$, $b$, and $c$ represent the hundreds, tens, and units digits, respectively. Then we have $N = 11^2a + 11b + c = 121a + 11b + c$.

We are also given that when $N$ is expressed in base $12$, it has the digits $cba$. This means $N = 12^2c + 12b + a = 144c + 12b + a$.

Setting these two expressions for $N$ equal, we have $121a + 11b + c = 144c + 12b + a$. This simplifies to $120a + 108c = 1b$. Since $b$ is the tens digit, it must be less than 9, so $b = 0$. Then our equation becomes $120a + 108c = 0$, which simplifies to $10a + 9c = 0$. The only way this can happen with $a$, $c$ being nonnegative integers is if $a = 9$ and $c = -10$. Since $c$ cannot be negative, we must have $c = 1$. Therefore, our number $N$ is $911_{11} = 1194_{12}$.

Therefore, $N = \boxed{1194}$.