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 $8$, it has the same three digits, in reverse order. What is $N$? (Express your answer in decimal.)

Answer 1

Let $N = abc_{11}$, where $a$, $b$, and $c$ are the hundreds, tens, and units digits, respectively, of $N$. Then, expressed in base $8$, $N = cba_8$.

Converting from base $11$ to base $10$, we have $N = 121a + 11b + c$. Converting from base $8$ to base $10$, we have $N = 64c + 8b + a$.

Setting these two expressions for $N$ equal, we have $121a + 11b + c = 64c + 8b + a$. Rearranging, we get $120a + 3b = 63c$.

Since $a$, $b$, and $c$ are digits in base $11$ and $8$, we know that $0 \leq a \leq 10$, $0 \leq b \leq 10$, and $0\leq c\leq 7$. From the equation $120a+3b=63c$, we know that $63c$ is divisible by $3$, so $c$ must be divisible by $3$. Thus, $c=0$, $c=3$, or $c=6$.

If $c=0$, then $120a+3b=0\cdot 63=0$, which implies that $a=b=0$. But $N$ has three digits, so $a\not=0$. Therefore, $c$ cannot be $0$.

If $c=3$, then $120a+3b=3\cdot 63=189$. Taking this equation modulo $3$, $120a\equiv 0\pmod{3}$, so $a\equiv 0\pmod{3}$. Then $a=3$ is the only acceptable value for $a$. We have $120\cdot3+3b=189$, or $b=3$. Thus, $N=333_{11}$, which is equal to $\boxed{255}$ in base $10$.

If $c=6$, then $120a+3b=6\cdot 63=378$. Taking this equation modulo $3$, $120a\equiv 0\pmod{3}$, so $a\equiv 0\pmod{3}$. Then $a=3$ is the only acceptable value for $a$. We have $120\cdot3+3b=378$, or $b=6$. Thus, $N=366_{11}$, which is not a three-digit number.

Therefore, the answer is $\boxed{255}$.