1. A student did not notice the multiplication sign between two seven-digit numbers and wrote a single fourteen-digit number, which turned out to be three times their product. Find these numbers.
1. A similar problem was in the 8th grade (problem 2). More detailed solutions are provided there.

First method. Let $x$ and $y$ be the required numbers, then we can form the equation: $3 \cdot x \cdot y = 10^7 x + y$ (see fact 11), $3 y = 10^7 + \frac{y}{x}$. Since $\frac{y}{x}$ is a number between 0 and 10, we have $10^7 < 3 y < 10^7 + 10$, hence $3333333 \frac{1}{3} < y < 3333336 \frac{2}{3}$. Now, we can either consider three cases: $y = 3333334$, $y = 3333335$, and $y = 3333336$, or notice that

$$
\frac{y}{x} \leq \frac{3333336}{1000000} < 4,
$$

which means $10^7 + 1 \leq 3 y < 10^7 + 4$. Only one number in this interval is divisible by 3, which is $10^7 + 2$. Therefore, $3 y = 10^7 + 2$, $y = 3333334$, $x = 1666667$.

Second method. $10^7 x = (3 x - 1) y$. Since $x$ and $(3 x - 1)$ have no common divisors, $3 x - 1$ is a divisor of $10^7$ (see facts 9 and 5). But $3 x - 1 \geq 3 \cdot 10^6 - 1$, so either $3 x - 1 = 5 \cdot 10^6$ or $3 x - 1 = 10^7$. Only $3 x - 1 = 5 \cdot 10^6$ fits, from which $x = 1666667$. From the first equation, we get: $y = 3333334$.