8. (4 points) The decimal representation of a four-digit number ABCD, consisting of different digits, was divided into three parts in three ways, after which three quadratic equations $\mathrm{A} x^{2}+B x+B \Gamma=0, \mathrm{~A} x^{2}+\mathrm{B} \mathrm{B} x+\Gamma=0$ and $\mathrm{A} x^{2}+\mathrm{B} x+\Gamma=0$ were formed. It turned out that all of them have roots. Find all possible values of ABCD.

AB, BC, and CD here are not products of digits, but two-digit numbers. Not only in the representation of the number ABCD, but also in the representations of these two-digit numbers, the first digits cannot be equal to 0.
Answer: 1920 or 2910.

Solution: First, we will prove that $\Gamma=0$. From the first equation, we get that $Б^{2} \geqslant 4 \cdot \mathrm{A} \cdot \mathrm{B}$. Since $\mathrm{Б}^{2} \leqslant 81$, we get that $\mathrm{B} \Gamma \leqslant 20$, which means $\Gamma=0$ or $\mathrm{B}=1$ (since zero cannot be equal to any digit other than $\Gamma$).

If $\mathrm{B}=1$, from the third equation we get $1=\mathrm{B}^{2}>4 \cdot \mathrm{AБ} \cdot \Gamma$, which is only possible when the right side is 0, i.e., when $\Gamma=0$. This condition ensures that the second and third equations have roots.

Returning to the first equation. ВГ $=10 \cdot \mathrm{B}$, from which $81 \geqslant \mathrm{Б}^{2} \geqslant 40 \cdot \mathrm{A} \cdot \mathrm{B}$. Since A and B are different digits, one of them is 1 and the other is 2. We also get that Б $^{2} \geqslant 80$, from which $\mathrm{E}=9$.

## Open Mathematics Olympiad for School Students

March 11, 2018

## 8th grade

## 2nd variant