On the cover of a 1958 issue of the German mathematical journal titled "Archimedes," the following numerical curiosity is visible: $1958-19 \cdot 58=8591-85 \cdot 91$. Is there another $\overline{a b c d}$ number with this property, that is, for which the following holds: $\overline{a b c d}-\overline{a b} \cdot \overline{c d}=\overline{d c b a}-\overline{d c} \cdot \overline{b a} ?$
I. Solution: We are looking for digits $a, b, c, d$, i.e., integers between 0 and 9, including both limits, such that

$$
\begin{aligned}
& 1000 a + 100 b + 10 c + d - (10 a + b)(10 c + d) = \\
& = 1000 d + 100 c + 10 b + a - (10 d + c)(10 b + a)
\end{aligned}
$$

Considering the commutativity of the two sides, we can require that $a \leq d$, and if $a = d$, then $b \leq c$. This prevents us from getting the solutions twice. By multiplying, rearranging, and dividing,

$$
111(a - d) + 10(b - c) + 11(b d - a c) = 0
$$

Notice that the first two coefficients on the left side are neighbors of numbers divisible by 11. Based on this, (1) can be rewritten as:

$$
11[10(a - d) + (b - c) + b d - a c] = (d - a) + (b - c)
$$

which means the right side is divisible by 11.

On the right side, $d - a$ cannot be negative, and its value can be at most 9:

$$
0 \leq d - a \leq 9
$$

Similarly,

$$
-9 \leq b - c \leq 9
$$

By adding these inequalities, the right side of (2) gives:

$$
-9 \leq (d - a) + (b - c) \leq 18
$$

Since this is divisible by 11, its value can only be 0 or 11. Therefore, rearranging,

$$
\begin{aligned}
\text { either I. } & c - b = d - a \\
\text { or II. } & c - b = d - a - 11
\end{aligned}
$$

Using the identity $b d - a c = (d - a) b - (c - b) a$, (2) can be rewritten so that each term contains either $d - a$ or $c - b$:

$$
11[(b - 10)(d - a) - (a + 1)(c - b)] = (d - a) + (b - c)
$$

In the I. case, simplifying by 11,

$$
[(b - 10) - (a + 1)](d - a) = (b - a - 11)(d - a) = 0
$$

This can only hold if $d - a = 0$, i.e., $d = a$, because the equation $b - a - 11 = 0$ cannot be satisfied given our notation. Thus, $c - b = 0$, i.e., $c = b$; we end up with the trivial result that every number of the form $\overline{a b b a}$ satisfies our requirement.

In the II. case, similarly, we get

$$
(b - a - 11)(d - a) + 11(a + 1) = 1
$$

or, since by assumption $b - a - 11 = c - d$,

$$
(d - c)(d - a) = 11a + 10
$$

Thus, the right side must be at least 10, i.e., positive, so both factors on the left side must be positive. Therefore, $a$ can be at most 8, and the right side can be one of the following:

$$
10, \quad 21, \quad 32, \quad 43, \quad 54, \quad 65, \quad 76, \quad 87
$$

However, 43, 65, 76, and 87 do not satisfy the condition of being factorizable into two positive integers less than 10. The value 54 = 6 * 9 does not work either, because if $a = 4$, then $d - a = 6$ or 9 implies $d \geq 10$, which is not a digit. Similarly, if the right side of (4) is 32 = 4 * 8, then $a = 2$, and only $d = 6$ works, but this leads to $c = -2$. Similarly, 21 = 3 * 7 and 10 = 2 * 5 also lead to negative $c$, so only the following two possibilities remain:

1) If $a = 1$, the right side of (4) is 21 = 3 * 7, and here $d - a = 7$ and $d - c = 3$ give $d = 8, c = 5$, and finally $b = 9$, leading to the example number 1958;
2) If $a = 0$, the right side of (4) is 10 = 2 * 5, and here $d - a = 5$ and $d - c = 2$ give $d = 5, c = 3, b = 9$. Thus,

$$
0935 - 09 \cdot 35 = 5390 - 53 \cdot 90 ( = 620 )
$$

However, due to the leading and thus unnecessary 0, this is only a degenerate solution.

Péter Tekulics (Szeged, Radnóti M. g. I. o. t.)

II. Solution: The equation (1) with the substitutions $b - c = x, b d - a c = y, d - a = e$ can be written as

$$
10 x + 11 y = 111 e
$$

Considering $x, y$ as unknowns and $e$ taking the values 0, 1, 2, ..., 9, $x$ must satisfy (3). Ignoring this constraint for now, a solution to the equation for every $e$ is $x_0 = -e, y_0 = 11e$. If $x, y$ is another integer solution, then

$$
10 x + 11 y = 111 e = 11(11 e) - 10 e
$$

Rearranging,

$$
10(x + e) = 11(11 e - y)
$$

Writing the left side as $11(x + e) - (x + e)$, it is only divisible by 11 if $x + e$ is divisible by 11: $x + e = 11k$, or $x = 11k - e$. Then from (6), $y = 11e - 10k$.

The inequality (3) for $x$ is only satisfied for $k = 0$, and for $e \geq 2$ also for $k = 1$.

If $k = 0$, then $b - c = x = -e$ and $d - a = e$ give $d = a + e$. Thus, $y = b d - a c = (c - a - e)e$. Equating this to $y = 11e$, if $e \neq 0$, we get $c - a = 11 + e \geq 12$, which cannot be satisfied by digits. If $e = 0$, then the conditions $b = c$ and $d = a$ give the trivial solution of the form $\overline{a b b a}$.

If $k = 1$, then $x = -e + 11, y = 11e - 10$. Similarly, $b = c - e + 11$ and $y = b d - a c = (c - e + 11)(a + e) - a c = (c - a - e)e + 11(a + e) = 11e - 10$. Rearranging,

| e | Thus the form of (7) | Solution |
| :---: | :---: | :---: |
| $\overline{4}$ | $4 c + 7 a = 6$ | No solution |
| 5 | $5 c + 6 a = 15$ | $c = 3, \quad a = 0 ; \quad d = 5, \quad b = 9$ |
| 6 | $6 c + 5 a = 26$ | $c = 1, \quad a = 4 ; \quad d = 10$ leads to no solution |
| 7 | $7 c + 4 a = 39$ | $\left\{\begin{array}{cc}c = 5 & a = 1, \quad d = 8, \quad b = 9 \\ c = 1, & a = 8 ; \quad d = 15 \text { leads to no solution }\end{array}\right.$ |
| 8 | $8 c + 3 a = 54$ | $c = 6, \quad a = 2 ; \quad d = 10$ leads to no solution |

$$
e c + (11 - e) a = e^2 - 10
$$

The left side cannot be negative, so only $e \geq 4$ can yield a solution