Junior olympiad question: Minimum value of 3 digit number divided by sum of its digits I recently had a maths competition where we were given this problem. I solved the question, but I narrowed down the possibilities then did more of a guess and check method. I was hoping someone else could help me get the answer to this question using a more efficient, less time consuming way. The answer that I got was 189/18 which was 10.5. I know it seems significantly easy but I just can't find the correct method for it. 
Let A be a number consisting of three different nonzero digits and let B be the sum of all the three digits. Find the minimum value of A/B.

Let the number be $\overline{abc}=100a+10b+c\,$ with digits $1 \le a,b,c \le 9\,$. Then:
$$
\begin{align}
\frac{100a+10b+c}{a+b+c} & = 1 + 9\cdot\frac{11a+b}{a+b+c} \\[3px]
& \ge 1 + 9\cdot\frac{11a+b}{a+b+\color{red}{9}} \quad\quad\quad\quad\text{(*)}\\[3px]
& = 1 + 9 + 9 \cdot \frac{10a-9}{a+b+9} \\[3px]
& \ge 1 + 9 + 9 \cdot \frac{10a-9}{a+\color{red}{8}+9} \\[3px]
& = 10 + 9 \cdot \frac{10a+170-179}{a+17} \\[3px]
& = 10 + 9 \cdot 10 - 9 \cdot \frac{179}{a+17} \\[3px]
& \ge 100 - 9 \cdot \frac{179}{\color{red}{1}+17} \\[3px]
& = \frac{189}{18}
\end{align}
$$
The minimum is attained when all the inequalities above are equalities i.e. $\overline{abc}=189\,$.
$(*)\,$ Given the condition that the digits must be different, and given that the next digit $b$ will also compete for the highest value, the alternative is to assign $c=8$ at this step, and save the digit $9$ to be assigned to $b=9$ at the next step. However, that leads to the solution $\overline{abc}=198\,$, which gives a higher ratio $198 / 18 \gt 189 / 18\,$, therefore the unique minimum is indeed attained for $\overline{abc}=189\,$.
