7.3. How many seven-digit natural numbers exist, for which the product of the first three digits is 30, the product of the three digits in the center is 7, and the product of the last three digits is 15?
Answer: 4.

Solution: Let the number be $\overline{\operatorname{abcdefg}}$. According to the condition, $c d e=7$, which means one of these digits is 7, and the other two are 1. Since 30 and 15 are not divisible by $7, d=7, c=e=1$. The number $30=5 \cdot 6 \cdot 1$, so we get two three-digit numbers $\overline{a b c}$: 561 and 651. The number $15=1 \cdot 3 \cdot 5$, from which we get two three-digit numbers $\overline{e f g}$: 135 and 153. Finally, we get $2 \cdot 2=4$ numbers.

Comment: Full solution of the problem - 7 points. Only the values of the three middle digits found - 2 points. Correct answer given without explanation - 0 points.