Let's determine the four-digit (natural) number $A B A C$, knowing that when multiplied by the three-digit numbers $A F G$, $C H B$, $F J F$, and $D E C$, we always get a product written with seven identical digits, and the digits in the products are $D$, $A$, $E$, and $J$. - What can we say about the decimal representation of the quotient $1: A B A C$ based on this?
I. Solution. I. Let us denote the number represented by 7 ones, $\left(10^{7}-1\right) / 9$, as $h$. Thus,

$$
\begin{gathered}
A B A C \cdot A F G=D D D D D D D=D \cdot h \\
A B A C \cdot C H B=A \cdot h \\
A B A C \cdot F J F=E \cdot h \\
A B A C \cdot D E C=J \cdot h
\end{gathered}
$$

By convention, we assume that different letters represent different digits, and no leading digit is 0. In the equalities (1)-(4), the digits $A, B, C, D, E, F, G, J$ cannot be 0 or 5. Indeed, $A, C, D, E, F, J$ as leading digits cannot be 0, and $B$ and $G$ cannot be 0 either, because then $A$ and $D$ would also be 0. The factors cannot end in 5 either, because otherwise either two different letters (a factor and the last digit of the product) would be the same (5), or the product would end in 0, but we have seen that $A, D, E, J$ are not 0; thus, the last digit of the product cannot be 5 either.

We will show that in the case of $AC$, the left side of (2) is $C H B A \cdot h = A \cdot \frac{10^{7}-1}{9} \\
9(A+1) \cdot 10^{5} \geqq 10^{7}, & A+1 \geqq 10^{2} / 9 > 11
\end{array}
$$

which is impossible, since $A \leqq 9$.

Therefore, the number in (2) is larger than the number in (1), so $A > D$. Similarly, from the comparison of (4) and (1), $J > A$. Indeed, if $B=1$, then $s$ would be a multiple of 7, which is impossible since 7 is a prime number. If $s=1$, then only one type of remainder could appear in the division, but the first two remainders of the division $1 / N$ are 1 and 10, which are different. Thus, the segment of the division $1 / N$ is seven digits long; from (4),

$$
\frac{1}{A B A C}=\frac{D E C}{J \cdot h}=\frac{239}{1111111}=\frac{2151}{9999999}=0, \dot{0} 00215 \dot{1}
$$

(Indeed, by subtracting the first segment, we get the $1 / 10^{7}$ part of the fraction:

$$
\left(\frac{2151}{9999999}-0,0002151=\frac{2151}{9999999} \cdot 10^{-7} .\right)
$$

Gyöngyi Barcza (Budapest, Apáczai Csere J. Gyak. G. III. o. t.) Zoltán Losonci (Szeged, Vedres I. Ép. Ip. T. III. o. t.)

II. Solution to the first question of the problem. We can quickly reach the goal if we know the following factorization of the number 1111111 from somewhere: $4649 \cdot 239$, and we also know that both factors are prime numbers. Thus, 4649 can only be a divisor of the factor $A B A C$, and the value of $A B A C$ is 4649 or 9298. The latter is impossible, because even its 110th multiple is an 8-digit number, so $A=4, B=6, C=9$. On the other hand, now the multipliers can only be 3-digit multiples of the 239 factor. There are only four such: 239, 478, 717, and 956. These can be reconciled with the already existing digits as follows: $239=D E C, 478=A F G$, $956=C H B$, and finally $F J F=717$, so all four equalities are satisfied and no other solution is possible.

Péter Szeredi (Budapest, Rákóczi F. G. III. o. t.)

Remark. After determining that $J=1$, we can search by trial and error for which value of $F$ the multiplier $F J F = F 1 F$ is a divisor of $h$ or a multiple of 10 less than $h$. For $F=2$, $212=4 \cdot 53$, so only $4h$ and $8h$ need to be tried, but divisibility does not hold; 313 is a prime number and is not a divisor of $h$, $414=2 \cdot 3^{2} \cdot 23$, but $2 \cdot 3^{2} h$ is already an 8-digit number; 515 is not possible; $616=11 \cdot 56$, but 11 is not a divisor of $h$, and $11h$ is already an 8-digit number. Finally, the trial with 717 quickly leads to the goal.