SOLUTION. Since $Y$ is a three-digit number, the five-digit number with the notation $X Y$ is the number $1000 X + Y$. Therefore, the student calculated the example $(1000 X + Y) : Z$ and according to the problem, the result was seven times greater than the original example, that is,

$$
\frac{1000 X + Y}{Z} = 7 \cdot \frac{X \cdot Y}{Z}
$$

From this, by multiplying by the number $Z$, we get the equation $1000 X + Y = 7 X Y$, which we solve for the unknown $Y$:

$$
Y = \frac{1000 X}{7 X - 1}
$$

For which $X$ is the last fraction an integer? In other words: when is the number $1000 X$ divisible by the number $7 X - 1$? Since the numbers $X$ and $7 X - 1$ are coprime (two consecutive numbers $7 X - 1$ and $7 X$ are coprime), we are looking for those $X$ for which the number $7 X - 1$ divides the number 1000. To avoid listing all divisors of the number 1000, we realize that $X$ is
two-digit, so $69 \leqq 7 X - 1 \leqq 692$. Therefore, let's factorize the number 1000 in all possible ways into the product of two factors such that one (say the first) of the factors is in the interval $\langle 69, 692 \rangle$:

$$
1000 = 500 \cdot 2 = 250 \cdot 4 = 200 \cdot 5 = 125 \cdot 8 = 100 \cdot 10 \text{.}
$$

From the equations

$$
7 X - 1 = 500, 7 X - 1 = 250, 7 X - 1 = 200, 7 X - 1 = 125, 7 X - 1 = 100
$$

only the equation $7 X - 1 = 125$ has an integer solution $X = 18$, for which we get $Y = 1000 X / (7 X - 1) = 1000 \cdot 18 / 125 = 144$.

Now we determine the unknown number $Z$. To do this, we use the condition of the problem that the value of the expression $X \cdot Y : Z$ is a natural number. Since $X = 18$ and $Y = 144$, it is the number $18 \cdot 144 : Z$, i.e., the number $2^{5} \cdot 3^{4} : Z$. Such a number is an integer if and only if the number $Z$ has a prime factorization of the form $2^{a} 3^{b}$, where $0 \leqq a \leqq 5$ and $0 \leqq b \leqq 4$. The exponents $a, b$ we find from the condition that the number $Z = 2^{a} 3^{b}$ is, according to the problem, a three-digit number and has the digit 2 in the units place. Since $3^{4} = 81$ and $2^{5} \cdot 3 = 96$, it must be $a \geqq 1$ and $b \geqq 2$. We now list all numbers $2^{a} 3^{b}$, where $a \in \{1, 2, 3, 4, 5\}$ and $b \in \{2, 3, 4\}$ in a table:

| $b$ | 1 | 2 | 3 | 4 | 5 |
| ---: | ---: | ---: | ---: | ---: | ---: |
| 2 | 18 | 36 | 72 | 144 | 288 |
| 3 | 54 | 108 | 216 | 432 | 864 |
| 4 | 162 | 324 | 648 | 1296 | 2592 |

Of the calculated numbers, only the numbers $Z = 432 = 2^{4} 3^{3}$ and $Z = 162 = 2^{1} 3^{4}$ have the required property.

Answer: The problem has two solutions. The student was supposed to calculate either the example $18 \cdot 144 : 432$ or the example $18 \cdot 144 : 162$.
3. The student was supposed to calculate the example $X \cdot Y: Z$, where $X$ is a two-digit number, $Y$ is a three-digit number, and $Z$ is a three-digit number with the digit 2 in the units place. The result of the example was supposed to be a natural number. However, the student overlooked the dot and interpreted the product $X \cdot Y$ as a five-digit number. As a result, he obtained a result seven times larger than it should have been. What example was the student supposed to calculate?