Task 1 - 231211

In the following schema (see figure), a digit is to be entered in each empty field so that a correctly calculated division problem results. In particular, none of the multi-digit numbers in the completed schema may have the digit 0 as the first digit.

Prove that there is exactly one entry of digits in the empty fields that meets these requirements! Determine this entry!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2143.jpg?height=512&width=714&top_left_y=435&top_left_x=1094)
If an entry of digits meets the requirements, and if the dividend is denoted by $x$, the divisor by $y$, and the quotient by $z$, then:

$$
\begin{gathered}
x: y=z \\
100 \leq y \leq 999
\end{gathered}
$$

Furthermore, if the number formed by the first four digits of the dividend is denoted by $t$, then: $9000 \leq$ $t \leq 9999$.

If the number in the second line of the schema, which is a multiple of $y$, is denoted by $n y$, where $1 \leq n \leq 9$, then:

$$
\begin{gathered}
t-n y \leq 9 \quad, \quad n \geq t-9 \geq 9000-9=8991 \quad \text { so } \\
y \geq \frac{8991}{n} \geq \frac{8991}{9}=999
\end{gathered}
$$

From (2) and (4), it follows that $y=999$. From (3), it further follows that

$$
n \geq \frac{8991}{y}=\frac{8991}{999}=9
$$

so $n=9$. Therefore, the number in the second line of the schema is $9 y=8991$, while in the 4th, 6th, 8th, and 10th lines of the schema, the number $y=999$ stands, because the only non-zero integer three-digit multiple of 999 is 999 itself. This leads to

$$
z=90100010011
$$

and it can only be the entry

$$
\begin{aligned}
& \begin{array}{ccccccccccccccccc}
9 & 0 & 0 & 0 & 9 & 9 & 1 & 0 & 0 & 0 & 0 & 9 & 8 & 9 & : & 999 & = \\
90100010011
\end{array} \\
& \begin{array}{cccccc}
8 & 9 & 9 & 1 & & \\
\hline & & & 9 & 9 & 9
\end{array} \\
& \begin{array}{llllllll}
9 & 9 & 9 & & & & \\
\hline & & & 1 & 0 & 0 & 0
\end{array} \\
& \begin{array}{cccccc}
9 & 9 & 9 & & & \\
\hline & & 1 & 0 & 9 & 8
\end{array} \\
& \begin{array}{llll}
9 & 9 & 9 & \\
\hline & 9 & 9 & 9
\end{array} \\
& \begin{array}{lrl}
9 & 9 & 9 \\
\hline & & 0
\end{array}
\end{aligned}
$$

This indeed represents a correctly solved division problem. Thus, it is proven that this entry is the only one that meets the given requirements.

Adapted from [5]