A student did not notice multiplication sign between two three-digit numbers and wrote it as a six-digit number. Result was 7 times more that it should be. Find these numbers.

[i](2 points)[/i]
1. Let the two three-digit numbers be \( x \) and \( y \). The student wrote the number as a six-digit number by concatenating \( x \) and \( y \), which can be expressed as \( 1000x + y \).

2. According to the problem, this six-digit number is 7 times more than the product of the two numbers. Therefore, we have the equation:
   \[
   1000x + y = 7xy
   \]

3. Rearrange the equation to isolate terms involving \( x \) and \( y \):
   \[
   1000x + y = 7xy \implies 1000x + y - 7xy = 0
   \]

4. Factor the equation:
   \[
   1000x + y - 7xy = 0 \implies y - 7xy = -1000x \implies y(1 - 7x) = -1000x
   \]

5. Solve for \( y \):
   \[
   y = \frac{1000x}{7x - 1}
   \]

6. Since \( x \) and \( y \) are three-digit numbers, \( y \) must also be an integer. Therefore, \( 7x - 1 \) must be a divisor of \( 1000x \).

7. The divisors of \( 1000 \) are \( 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000 \). We need to find which of these divisors, when added to 1, are divisible by 7.

8. Check each divisor:
   - \( 1 + 1 = 2 \) (not divisible by 7)
   - \( 2 + 1 = 3 \) (not divisible by 7)
   - \( 4 + 1 = 5 \) (not divisible by 7)
   - \( 5 + 1 = 6 \) (not divisible by 7)
   - \( 8 + 1 = 9 \) (not divisible by 7)
   - \( 10 + 1 = 11 \) (not divisible by 7)
   - \( 20 + 1 = 21 \) (divisible by 7)
   - \( 25 + 1 = 26 \) (not divisible by 7)
   - \( 40 + 1 = 41 \) (not divisible by 7)
   - \( 50 + 1 = 51 \) (not divisible by 7)
   - \( 100 + 1 = 101 \) (not divisible by 7)
   - \( 125 + 1 = 126 \) (divisible by 7)
   - \( 200 + 1 = 201 \) (not divisible by 7)
   - \( 250 + 1 = 251 \) (not divisible by 7)
   - \( 500 + 1 = 501 \) (not divisible by 7)
   - \( 1000 + 1 = 1001 \) (divisible by 7)

9. The valid divisors are \( 20, 125, \) and \( 1000 \). We now solve for \( x \) and \( y \) using these divisors:
   - For \( 7x - 1 = 20 \):
     \[
     7x = 21 \implies x = 3 \implies y = \frac{1000 \cdot 3}{20} = 150
     \]
   - For \( 7x - 1 = 125 \):
     \[
     7x = 126 \implies x = 18 \implies y = \frac{1000 \cdot 18}{125} = 144
     \]
   - For \( 7x - 1 = 1000 \):
     \[
     7x = 1001 \implies x = 143 \implies y = \frac{1000 \cdot 143}{1000} = 143
     \]

10. The possible pairs \((x, y)\) are \((3, 150)\), \((18, 144)\), and \((143, 143)\).

The final answer is \(\boxed{(143, 143)}\).