D is a 3-digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred's place and the digit at the unit's place of D ?

• Let the 3-digit number be $100a + 10b + c$, where $a \in \{1, \dots, 9\}$ and $b, c \in \{0, \dots, 9\}$.

• The ratio is $R = \frac{100a + 10b + c}{a + b + c}$. To minimize $R$, we need to minimize the numerator relative to the denominator.

• Since $100a$ dominates the numerator, we should choose the smallest possible $a$, which is $a=1$.

• With $a=1$, $R = \frac{100 + 10b + c}{1 + b + c}$. To minimize this, we need to maximize the denominator $(1+b+c)$ while keeping the numerator small. This suggests making $b$ and $c$ large.

• Let's test the maximum values for $b$ and $c$. If $b=9$ and $c=9$, the number is $199$. The sum of digits is $1+9+9=19$. The ratio is $199/19 \approx 10.47$.

• Let's check if a smaller ratio is possible. If we decrease $c$ to 8, number is $198$, sum is $18$, ratio $198/18 = 11$. If we decrease $b$ to 8, number is $189$, sum is $18$, ratio $189/18 = 10.5$. If we decrease both, e.g., $188$, sum $17$, ratio $188/17 \approx 11.05$.

• Comparing $199/19 \approx 10.47$ with other candidates like $198/18=11$, $189/18=10.5$, $188/17 \approx 11.06$, the minimum ratio occurs at $D=199$.

• For $D=199$, the digit at the hundred's place is $1$ and the digit at the unit's place is $9$. The difference is $|1 - 9| = 8$.