Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$. 
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$.

$m$ is a 3 digit number (because this was an AIME problem). 
$$m \equiv 0 \pmod{17}$$
$$m \equiv 17 \pmod{9} \equiv -1 \pmod{9}$$
Applying the chinese remainder theorem, the solution is supposed to be:
$$m = 17 \cdot 9 = 153$$
but that isnt correct. 

$\begin{cases}m\equiv 0\equiv 17\pmod {17}\\m\equiv 17\pmod{9}\end{cases}\implies 17, 9\mid m-17\stackrel{(17,9)=1}\implies 153\mid m-17$   
So necessarily $m\in\{17,170,323,476,\ldots\}$. After checking $17,170,323,476$ in that order, only $476$ satisfies the sufficient conditions (of digits summing up to $17$). Answer: $476$.   
This uses the hint by Bill Dubuque in the comments to op, so this answer is community wiki.
