## Task 2 - 271212

Determine all two-digit and all three-digit natural numbers for which the product of the digits is twice as large as the cross sum!
I. If a two-digit number $z$ has the required property, it follows:

The digits of $z$ are natural numbers $a, b$, for which $00$; hence and because of (3), $b>0$ and $c>0$ must also be true, so that in addition to (2), 

$$
0<a, b, c \leq 9
$$

holds. Because of (3), at least one of the numbers $a, b, c$ must be even. Since (3) and (2') remain valid for any permutation of $a, b, c$, it suffices to consider the case where $c$ is even, i.e., $c$ is one of the numbers 2, 4, 6, 8.

The following table contains for these values and all $b=1,2, \ldots, 9$ each time the equation (4) divided by 2 and information about its solution $a$:

| $\mathrm{b}$ | $c=2$ |  | $c=4$ |  | $c=6$ |  |  |  |
| :--- | :--- | :---: | :--- | :---: | :--- | :--- | :--- | :---: |
| 1 | $0 a=3$ | - | $1 a=5$ | $a=5$ | $2 a=7$ | - | $3 a=9$ | $a=3$ |
| 2 | $1 a=4$ | $a=4$ | $3 a=6$ | $a=2$ | $5 a=9$ | - | $7 a=10$ | - |
| 3 | $2 a=5$ | - | $5 a=7$ | - | $8 a=9$ | - | $11 a=11$ | $a=1$ |
| 4 | $3 a=6$ | $a=2$ | $7 a=8$ | - | $11 a=10$ | - | $15 a=12$ | - |
| 5 | $4 a=7$ | - | $9 a=9$ | $a=1$ | $14 a=11$ | - | $19 a=13$ | - |
| 6 | $5 a=8$ | - | $11 a=10$ | - | $17 a=12$ | - | $23 a=14$ | - |
| 7 | $6 a=9$ | - | $13 a=11$ | - | $20 a=13$ | - | $27 a=15$ | - |
| 8 | $7 a=10$ | - | $15 a=12$ | - | $23 a=14$ | - | $31 a=16$ | - |
| 9 | $8 a=11$ | - | $17 a=13$ | - | $26 a=15$ | - | $35 a=17$ | - |

From this, it follows that $z$ is one of the following numbers or a number derived from these by rearranging the digits: 422, 242, 514, 224, 154, 318, 138.

IV. If $z$ is such a number, then $z$ has the required property because $2 \cdot 2+4=16=2(2+2+24)$ or analogously for the other numbers.

With I., II., III., IV., it is proven that exactly the numbers 36, 44, 63, 138, 145, 154, 183, 224, 242, 318, 381, 422, 514, 541, 813, 831 have the required property.

Adapted from $[5]$