(The 3rd problem of the general competition:)

Determine the number $A B C C$ (written in the decimal system) if

$$
A B C C=(D D-E) \cdot 100+D D \cdot E
$$

where $A, B, C, D$ and $E$ are distinct digits.
The multi-digit numbers are transformed as follows:

$$
A B C C=1000 \cdot A+100 \cdot B+11 \cdot C, \quad D D=11 \cdot D
$$

Substituting these, we rearrange so that the terms with a factor of 100 are on the left side, and the others on the right side:

$$
(10 \cdot A+B-11 \cdot D+E) \cdot 100-11 \cdot(D \cdot E-C)
$$

Thus, the right side must also be divisible by 100. This is only possible if $D \cdot E-C$ is divisible by 100. Since $D$, $E$, and $C$ are all between 0 and 9 and are distinct, $D \cdot E-C$ is at most $9 \cdot 8-0=72$, and at least $0 \cdot 1-9=-9$. Between -9 and 72, there is only one number divisible by 100, which is 0, so $D \cdot E=C$. Furthermore, the left side of (2) must also be 0, so the expression in parentheses must be 0. We can write $10 \cdot A$ as $11 \cdot A-A$ and rearrange the equation:

$$
B+E-A=11 \cdot(D-A)
$$

Since $D$ and $A$ are distinct, the right side, and thus the left side, is a non-zero number. Moreover, the right side is divisible by 11, so the left side must also be. Therefore, $B+E-A$ is a non-zero (integer) number divisible by 11. On the other hand, since $B$, $A$, and $E$ are also between 0 and 9 and $A \neq 0$, $B+E-A$ is at most $9+8-1=16$, and at least $0+1-9=-8$. Between -8 and 16, there is only one non-zero number divisible by 11, which is 11. Therefore, $B+E-A=11$, from which $D-A=1$ follows.

So far, we have established the following:

$$
D \cdot E=C, \quad B+E-A=11 \quad \text{ and } \quad D=A+1
$$

From the third equality, solving for $A$ and substituting into the second:

$$
B+E=D+10
$$

Rearranging:

$$
E=D+(10-B)
$$

Since $B$ is at most 9, $10-B$ is at least 1, and thus $E$ is at least $D+1$, meaning $E$ is greater than $D$. Based on this, $D \cdot D$ is less than $D \cdot E$, or $C$, which is at most 9; that is, $D^2$ is less than $3^2$, so $D$ is less than 3. On the other hand, $D=A+1 \geq 2$, so $D=2$, from which $A=1$, and furthermore

$$
C=2E \quad \text{ and } \quad B+E=12
$$

Since $E>D$, $E \geq 3$. On the other hand, $E \leq 4$, otherwise $C$ would be a two-digit number. In the case $E=4$, $C=B(=8)$ results, which again does not satisfy the distinctness of the digits. The only remaining possibility is $E=3$, from which $C=6$, and $B=9$.

The values of the digits thus obtained indeed satisfy the conditions:

$$
1966=(22-3) \cdot 100+22 \cdot 3
$$