1. Determine the natural number $\overline{a b c d}$ such that for its digits the following conditions hold:

$$
a \cdot d + c \cdot d = 72, \quad a \cdot c + c \cdot d = 56 \text{ and } a \cdot b \cdot c \cdot d = 0
$$
## First Solution.

From the condition $a \cdot b \cdot c \cdot d=0$, it follows that at least one of the factors is equal to 0. If $d=0$, then the first equation would read $0=72$, which is impossible. If $c=0$, then the second given equation would read $0=56$, which is also impossible. If $a=0$, then $c \cdot d=72$ and $c \cdot d=56$ would both hold, which is also impossible. This means that it is necessary that $b=0$.

Let's write the expressions on the left sides of the first two given equations in the form of a product:

$$
d \cdot(a+c)=72 \quad \text{and} \quad c \cdot(a+d)=56
$$

Let's list all possibilities for $c$, $a$, and $d$ that satisfy the second equation, and check which of these possibilities also satisfy the first equation.

| $c$ | $a+d$ | $a$ | $d$ | $d \cdot(a+c)$ |
| :---: | :---: | :---: | :---: | :---: |
| 1 | 56 | $/$ | $/$ | $/$ |
| 2 | 28 | $/$ | $/$ | $/$ |
| 4 | 14 | 5 | 9 | 81 |
| 4 | 14 | 6 | 8 | 80 |
| 4 | 14 | 7 | 7 | 77 |
| 4 | 14 | 8 | 6 | 72 |
| 4 | 14 | 9 | 5 | 65 |
| 7 | 8 | 1 | 7 | 56 |
| 7 | 8 | 2 | 6 | 54 |
| 7 | 8 | 3 | 5 | 50 |


| $c$ | $a+d$ | $a$ | $d$ | $d \cdot(a+c)$ |
| :---: | :---: | :---: | :---: | :---: |
| 7 | 8 | 4 | 4 | 44 |
| 7 | 8 | 5 | 3 | 36 |
| 7 | 8 | 6 | 2 | 26 |
| 7 | 8 | 7 | 1 | 14 |
| 8 | 7 | 1 | 6 | 54 |
| 8 | 7 | 2 | 5 | 50 |
| 8 | 7 | 3 | 4 | 44 |
| 8 | 7 | 4 | 3 | 36 |
| 8 | 7 | 5 | 2 | 26 |
| 8 | 7 | 6 | 1 | 14 |

The only possibility that satisfies both conditions is $a=8, c=4$, and $d=6$, i.e., the desired number is 8046.