Find such a three-digit number, which has the smallest possible ratio of itself to its sum of digits?

1. Let’s denote the required three-digit number:

x = abc, where a, b and c are three-digit numbers;
x = 100a + 10b + c;
y = a + b + c, the sum of the digits of the number x.
2. For the ratios of these numbers, we get:

k = y / x = (100a + 10b + c) / (a + b + c).

Let’s select the whole part of the fraction:

k = (99a + 9b + a + b + c) / (a + b + c);
k = 1 + (99a + 9b) / (a + b + c);
k = 1 + 9 (11a + b) / (a + b + c). (one)
3. From equation (1) it follows that the smallest value for k is obtained with the largest values of b and c, and with the smallest value for ‘a’:

c = 9;
b = 9;
a = 1;
x = abc = 199.
k = 199/19 ≈ 10.47.
Answer: 199.