415. Five numbers were written on the board. Then these numbers were erased and their pairwise sums were written:

$$
4,8,10,12,14,18,19,21,25,29
$$

What five numbers were written on the board?

![](https://cdn.mathpix.com/cropped/2024_05_21_0aa0e6ac168e80891953g-127.jpg?height=586&width=1353&top_left_y=264&top_left_x=340)
Let's denote the sought numbers in ascending order as $x_{1}, x_{2}, x_{3}, x_{4}$, and $x_{5}:$

$$
x_{1}<x_{2}<x_{3}<x_{4}<x_{5}
$$

All inequalities here are strict, since there are no equal sums among the pairwise sums.

The problem is that we do not know which numbers, when added, result in a sum of, for example, 10, or a sum of 12.

The smallest of all pairwise sums is $x_{1}+x_{2}$. Indeed, if

$$
x_{1} \leq x_{i}(i=1,2,3,4,5), x_{2} \leq x_{k}(k=2,3,4,5)
$$

then, adding these inequalities term by term, we get: $x_{1}+x_{2} \leq x_{i}+x_{k}$.

Similarly, it can be shown that the next largest sum is $x_{1}+x_{3}$.

Further, the largest of all pairwise sums is $x_{4}+x_{5}$, and the one preceding it in magnitude is $x_{3}+x_{5}$.

We obtain a system of four equations with five unknowns. Where do we get the fifth equation? It is unclear, for example, which of the sums $x_{1}+x_{4}$ and $x_{2}+x_{3}$ is larger; in general, it can be different.

If we add all 10 pairwise sums, the new sum will be four times the sum of all five sought numbers (think about why). We have:

$$
\frac{1}{4}(4+8+10+12+14+18+19+21+25+29)=40
$$

It remains to solve the system of linear equations

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=4 \\
x_{1}+x_{3}=8 \\
x_{4}+x_{5}=29 \\
x_{3}+x_{5}=25 \\
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=40
\end{array}\right.
$$

In addition, the obtained solution needs to be verified by the pairwise sums that did not enter the system.

Answer: $1,3,7,11,18$.