7. Dad was driving his son to a math club lesson. On the way, they were overtaken by 4 cars. The son admitted to his father that he could only remember the numbers of the two smallest cars - 119 and 179, and that the others definitely had the digit 3. The father, however, remembered all of them and gave his son the following problem. If the numbers are paired and each pair is multiplied, and then the two resulting numbers are added, the result is 105080. Help the son remember the two remaining numbers. (All car numbers are three-digit numbers).
Solution: Let x and y be the numbers of the two remaining cars. Then there are two possible cases.

1st case. $\quad x y+119 \cdot 179=105080 \Rightarrow x y=83779=199 \cdot 421-\quad$ this $\quad$ is $\quad$ two $\quad$ prime factors, so there are no other three-digit factors. But in this case, there is no digit 3.

2nd case. $119 x+179 y=105080$ - this is a Diophantine equation. Let's solve it.

First, find any integer solution to the equation $119 x+179 y=1$. This is $x=-3, y=2$. Therefore, one solution to the original equation is $x=-315240$, $y=210160$, and the general solution is $x=179 t-315240, y=210160-119 t$, where $t \in Z$. Choose $t \in Z$ so that x and y become three-digit natural numbers. This will be when $t=1762,1763,1764,1765$. Four pairs of three-digit numbers are obtained: $x=158, y=482 ; x=337, y=363 ; x=516, y=244 ; x=695, y=125$. Only one variant satisfies the condition of the problem.

Answer: $337,363$.