3. Find all four-digit numbers $\overline{a b c d}$ that are divisible by each of the two-digit numbers $\overline{a b}, \overline{b c}, \overline{c d}$, where the digits $a, b, c, d$ are odd and not all the same.

The school - written part of the first round in category A takes place

on Tuesday, December 5, 2000

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
3. From the expression $\overline{a b c d}=100 \cdot \overline{a b}+\overline{c d}$, it follows that the divisibility conditions for the numbers $\overline{a b}$ and $\overline{c d}$ are satisfied if and only if $\overline{c d} \mid 100 \cdot \overline{a b}$ and $\overline{a b} \mid \overline{c d}$, i.e., if and only if $\overline{c d}=k \cdot \overline{a b}$, where the natural number $k$ is a divisor of the number 100. Since both numbers $\overline{a b}$ and $\overline{c d}$ are odd and two-digit, it follows that either $k=1$ or $k=5$. We will distinguish both cases, and with respect to the expression $\overline{a b c d}=10 \cdot \overline{b c}+(1000 a+d)$, we will instead of the condition $\overline{b c} \mid \overline{a b c d}$ examine the equivalent condition

$$
\overline{b c} \mid(1000 a+d) .
$$

(This modification is not necessary, it just simplifies the subsequent notation somewhat.)

a) If $\overline{c d}=\overline{a b}$, then $c=a$ and $d=b$, so the condition (P) can be written in the form $(10 b+a) \mid(1000 a+b)$. Since

$$
1000 \cdot(10 b+a)-(1000 a+b)=9999 \cdot b=11 \cdot 9 \cdot 101 \cdot b
$$

and 101 is a prime number (hence it is coprime with the number $10 b+a$), we get the equivalent condition $(10 b+a) \mid(11 \cdot 9 b)$. From this, considering the obvious inequality $10 b+a>9 b$, it follows that the number $10 b+a$ has the number 11 in its prime factorization. The condition $11 \mid(10 b+a)$ is satisfied only if the digits $a$ and $b$ are equal, but then from the equality $\overline{c d}=\overline{a b}$ it would follow that the number $\overline{a b c d}$ has all digits the same. According to the problem statement, we do not consider such numbers.

b) From the equality $\overline{c d}=5 \cdot \overline{a b}$, we immediately determine (odd) digits $a=1$ and $d=5$, after substituting them, we get the equality $b=2 c-9$, so there are three possibilities:

$$
c=5 \text { and } b=1, \quad c=7 \text { and } b=5, \quad c=9 \text { and } b=9 .
$$

The condition (P) now has the form $\overline{b c} \mid 1005$, but of the numbers 15, 57, and 99, only the number 15 is a divisor of the number $1005 (1005=3 \cdot 5 \cdot 67)$, so necessarily $b=1$ and $c=5$.

Answer: The sought number is unique, and it is 1155.

For a complete solution, 6 points are awarded, of which 3 points for justifying that either $\overline{c d}=\overline{a b}$ or $\overline{c d}=5 \cdot \overline{a b}$, 2 points for analyzing the first possibility, and 1 point for analyzing the second possibility.