In the standing multiplication, each letter and dot represents a digit, equal letters represent the same digit, different letters represent different digits. Let's write the appropriate digits in place of the letters and dots so that the multiplication is correct.

$$
\begin{aligned}
& a b b a \cdot c d e a \\
& \text {. . . } \\
& \text {. . . } d \text {. } \\
& \begin{array}{r}
\qquad b b e \\
. . c e e
\end{array}
\end{aligned}
$$
In the thousand's place of the fourth partial product, $a \cdot a=a^{2}=e$ has no remainder, so there is none in the unit's place, thus $e \leq 9$, $a \leq 3$. But $a \neq 0,1$, because in this case $a^{2}=a$, but $e$ differs from $a$. Similarly, there can be no remainder in the second and third place when multiplying by $a$, and thus $a \cdot b=b,(a-1) b=0$, which can only be satisfied with $b=0$ due to $a \neq 1$. The assumption $a=2$ is ruled out because then $e=4$, and the third partial product $2002 \cdot 4=8008$ would be four digits, whereas it must be five digits. Therefore, $a=3, e=9$. The third partial product $3003 \cdot 9=27027$, so on the one hand $d=2$, and on the other hand from the ten's place of the product $7+0=c$ we get $c=7$.

Accordingly, the product in question is $3003 \cdot 7293=21900879$, in which the first two partial products are five and four digits, respectively, so the remaining requirements are also met.

Rezső Gáspár (Debrecen, Kossuth L. Gyak. G. I. o. t.)