122. Four different three-digit integers, starting with the same digit, have the property that their sum is divisible by three of them without a remainder. Find these numbers.
122. Answer: $108,135,180,117$.

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the unknown numbers, $S$ be their sum, and $a$ be the first digit of each of them. It is clear that $100 a \leqslant x_{i}<100(a+1)$ $(i=1,2,3,4)$.

Using these inequalities, we get $x_{t}+300 a \leqslant S<$ l $<x_{i}+300(a+1)$, from which

$$
1+\frac{300 a}{x_{i}} \leqslant \frac{S}{x_{i}}<1+\frac{300(a+1)}{x_{i}}
$$

therefore,

$$
1+\frac{3 a}{a+1}<\frac{S}{x_{l}}<4+\frac{3}{a}
$$

For $a=1$, we get the inequalities $2.5<S / x_{i}<7$, and for $a \geqslant 2$: $3<S / x_{l}<5.5$. It remains to do a search.

Since three of the four quotients $S / x_{i}$ are different integers, the case $a \geqslant 2$ is impossible (there are only two integers between 3 and 5.5). Therefore, $a=1$ and the integer quotients should be sought among the numbers $3,4,5$, and 6. But 3 and 6 cannot both be quotients at the same time, since the ratio of any two three-digit numbers from 100 to 199 is less than two. Two possibilities remain: 1) three quotients $3,4,5$ and 2) three quotients $6,5,4$. In both cases, $S$ is divisible by $60: S=60 \mathrm{k}$. In the first case, the sought numbers are: $12 k, 15 k, 20 k$, and $13 k$.

The first digit is the same for all numbers only when $k=9$, which gives the answer. In the second case, the set of numbers $10 k, 12 k, 15 k$

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-151.jpg?height=262&width=310&top_left_y=1275&top_left_x=134)

Figure 60 and $23 k$ does not satisfy the condition of the problem for any $k$, since the ratio of the numbers $23 k$ and $10 k$ is greater than two.