Harvard-MIT November Tournament

Solution:
We assume that the answer is at most three digits (in base $10$). Then our desired number can be expressed in the form $\overline{a b c}_{10} = \overline{d e f}_{17}$, where $a, b, c$ are digits in base $10$, and $d, e, f$ are digits in base $17$. These variables then satisfy the equations
$$\begin{aligned} 100 a + 10 b + c &= 289 d + 17 e + f \\ a + b + c &= d + e + f \end{aligned}$$
Subtracting the second equation from the first, we obtain $99 a + 9 b = 288 d + 16 e$, or $9(11 a + b) = 16(18 d + e)$. From this equation, we find that $11 a + b$ must be divisible by $16$, and $18 d + e$ must be divisible by $9$.

To minimize $\overline{a b c}$, we find the minimal possible value of $a$:

If $a = 0$, then the only way for $11 a + b = b$ to be divisible by $16$ is to set $b = 0$; however, this is disallowed by the problem condition, which stipulates that the number must be greater than $10$.

If we try $a = 1$, then we find that the only possible value of $b$ which lets $11 a + b = b + 11$ be divisible by $16$ is $b = 5$. Plugging these in and simplifying, we find that we must have $18 d + e = 9$. The only possible solution to this is $d = 0, e = 9$.

Now to satisfy $a + b + c = d + e + f$, we must have $1 + 5 + c = 0 + 9 + f$, or $c = f + 3$. The minimal possible solution to this is $c = 3, f = 0$.

So our answer is $\overline{a b c} = 153$, which is also equal to $090_{17}$.