1. Find all 3-digit natural numbers $n$ with non-zero digits, for which it holds: if we remove the leftmost digit from the number $n$, we get a number that divides $n$.
II/1. Let's write $n=\overline{a b c}$. The condition of the problem states that $\overline{b c}$ divides $\overline{a b c}=100 a+\overline{b c}$, so it also divides $100a$. Suppose first that $\overline{b c}$ is not divisible by 5. Then $\overline{b c}$ must divide $4a$. Since $\overline{b c}$ is a two-digit number, $4a$ must also be a two-digit number, so $3 \leq a \leq 9$. Therefore, $4a$ can take the values $12, 16, 20, 24, 28, 32$, or 36. The two-digit divisors of these numbers are, respectively, 12, 16, 20 and 10, 24 and 12, 28 and 14, 32 and 16, and 36 and 18 and 12. Since the digits must be non-zero, we get the solutions $312, 416, 624, 612$, $728, 714, 832, 816, 936, 918$, and 912. If $\overline{b c}$ is divisible by 5, then $c=5$, since $c$ must be a non-zero digit. Therefore, $\overline{b c}=10 b+5=5(2 b+1)$, and thus $(2 b+1)$ must divide $20a$. Since $2 b+1$ is an odd number, it must divide $5a$. If $2 b+1$ is divisible by 5, then either $b=2$ (so $2 b+1=5$) and $a$ can be any value, or $b=7$ (so $2 b+1=15$) and $a$ can be 3, 6, or 9. This gives us the solutions $125, 225, 325, 425, 525, 625, 725, 825$, and 925, as well as 375, 675, and 975. If $2 b+1 \neq 5$, then it is coprime with 5 and must divide $a$. Therefore, $2 b+1 \leq 9$ and thus $b \leq 4$ and $b \neq 2$ (otherwise $2 b+1=5$). If $b$ is sequentially equal to 4, 3, and 1, then $a$ can be sequentially equal to 9, 7, and 3 or 6 or 9. This gives us the additional solutions $945, 735, 315, 615$, and 915.

In summary, all solutions are $125, 225, 312, 315, 325, 375, 416, 425, 525, 612, 615, 624, 625$, $675, 714, 725, 728, 735, 816, 825, 832, 912, 915, 918, 925, 936, 945, 975$.

\section*{Observation that $\overline{b c}$ divides $100a$ 1 point}

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-08.jpg?height=52&width=1633&top_left_y=1182&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-08.jpg?height=55&width=1633&top_left_y=1229&top_left_x=220)  Listing 27 or 28 solutions .................................................................. 4 points  If the contestant lists fewer than 27 solutions, 1 point is awarded for every 7 correct solutions. If the contestant lists numbers that do NOT SATISFY the conditions of the problem, 1 point is deducted for EACH INCORRECT SOLUTION.