2. A student did not notice the multiplication sign between two three-digit numbers and wrote a single six-digit number, which turned out to be seven times greater than their product. Find these numbers.
2. First method. Let $x, y$ be the desired three-digit numbers. If we append three zeros to the number $x$, we get the number $1000 x$, and if we append $y$, we get $1000 x + y$ (see fact 11).

Thus, the student wrote the number $1000 x + y$. According to the problem, this number is seven times greater than $x \cdot y$. This gives us the equation

$$
7 x \cdot y = 1000 x + y
$$

Divide both sides of equation (1) by $x$:

$$
7 y = 1000 + \frac{y}{x}
$$

The number $\frac{y}{x}$ is positive and less than 10, since $y \leqslant 999$ and $x \geqslant 100$. Therefore,

$$
1000 < 7 y < 1010
$$

Dividing this inequality by 7, we get

$$
142 \frac{6}{7} < y < 144 \frac{2}{7}
$$

Since $y$ is an integer, $y$ is either 143 or 144. Let $y = 143$. Substituting this value of $y$ into equation (1), we get:

$$
7 x \cdot 143 = 1000 x + 143
$$

Solving this equation, we find $x = 143$. If $y = 144$, then the analogous equation gives $x = 18$, which is not valid because $x$ is a three-digit number.

In the second method. Rewrite equation (1) as $1000 x = (7 x - 1) y$. It is easy to see that $x$ and $7 x - 1$ have no common divisors other than 1 and -1. Indeed, if $d$ is a common divisor of the numbers $x$ and $7 x - 1$, then $d$ is a divisor of the number $7 x$, and therefore a divisor of the number $1 = 7 x - (7 x - 1)$ (see fact 5). But 1 is divisible only by 1 and -1.

Thus, the number $7 x - 1$ is a divisor of the product $1000 \cdot x$ and is coprime with the second factor. Then, by a known theorem (see fact 9), the number $7 x - 1$ is a divisor of the number 1000. But

$$
7 x - 1 \geqslant 7 \cdot 100 - 1 = 699
$$

so $7 x - 1 = 1000$ (the only divisor of the number 1000 that is greater than or equal to 699 is the number 1000 itself), from which $x = 143$. Substituting $x = 143$ into the original equation, we find $y = 143$.

Comment. Compare with problem 1 for 10th grade.

![](https://cdn.mathpix.com/cropped/2024_05_21_6455f48598646f41a890g-126.jpg?height=410&width=523&top_left_y=189&top_left_x=126)

Fig. 41