Answer: 73.

Solution. Let the original numbers be denoted by $a$ and $b$. Then the specified seven-digit number will have the form $10000a + b$. According to the condition, $10000a + b = nab$, from which we get $b = \frac{10000}{na - 1}$.

Note that the numbers $a$ and $a-1$ do not have common divisors, so $na - 1 = p$ is a divisor of 10000. Moreover, $nab \geq 10000a$, which means $nb \geq 10000$, and $n \geq 10000 / 999 > 10$. Therefore, $n \geq 11$, $p \geq 1099$, so we need to check the values of $p$ equal to $10000, 5000, 2500, 2000, 1250$ and find those for which $p+1$ can be factored into a two-digit ($n$) and a three-digit ($a$) divisor.

Let $p = 10000$, then $10001 = 73 \cdot 137$, so $n = 73$, $a = 137$, $b = 10000 \cdot a / p = a$.

Let $p = 5000$, then $5001 = 3 \cdot 1667$, the number 1667 is prime; there are no three-digit divisors.

Let $p = 2500$, then $2501 = 41 \cdot 61$, there are no three-digit divisors.

Let $p = 2000$, then $2001 = 3 \cdot 23 \cdot 29$. Since $n \geq 11$, the suitable value is at least 23, but then $a \leq 3 \cdot 29 = 87$, which is a two-digit number.

Let $p = 1250$, then $1251 = 3 \cdot 417$, the number 417 is prime. There is no factorization into a two-digit and a three-digit number.

Thus, only the case $n = 73$ remains. Indeed, $1370137 / 73 = 18769 = 137 \cdot 137$.

Criteria. Only the answer - 1 point, the equation is correctly set up - 3 points. The correct answer is derived, but the enumeration is incomplete or unproven - 5 points.
9-5. Two three-digit numbers are written on the board in a multiplication example. If the multiplication sign is replaced with 0, a seven-digit number is obtained, which is an integer multiple of the product. By what factor exactly