# Task 1. (10 points)

Five different weights, each weighing a whole number of kilograms, were weighed in all possible groups of three weights. As a result, the following weights (in kilograms) of the ten weighed groups were obtained: $10,14,15,16,17,17,18$, $21,22,24$. Find the weights of these five weights.
Solution.

Let $a>b>c>d>e$ be the weights of the given weights. Since each weight is included in six out of ten weighed triples, the sum of the obtained weights is $6(a+b+c+d+e)$, therefore,

$$
a+b+c+d+e=\frac{10+14+15+16+17+17+18+21+22+24}{6}=29
$$

Meanwhile,

$$
\begin{aligned}
& a+b+c=24, c+d+e=10 \Rightarrow a+b+2 c+d+e=34 \Rightarrow c=34-29=5 \\
& a+b+d=22, c+d+e=10 \Rightarrow a+b+c+2 d+e=32 \Rightarrow d=32-29=3 \\
& a+b+c=24, b+d+e=14 \Rightarrow a+2 b+2 c+d+e=38 \Rightarrow b=38-29=9
\end{aligned}
$$

We obtain

$$
a+b=19 \Rightarrow a=10 \text { and } d+e=5 \Rightarrow e=2
$$

Answer. $10,9,5,3,2$.

| Content of the criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. |  |  |
| All main logical steps of the solution are provided.  The wrong answer is obtained due to a computational error or typo. | $\pm$ | 7 |
| Strict justification of some conclusions is missing. The wrong answer is obtained due to a computational error or typo. | $+/ 2$ | 5 |
| The answer is correct.  Strict justification of some conclusions is missing. In | $+/ 2$ | 5 |

All-Russian School Olympiad "Mission Possible. Your Calling - Financier!"

| particular, it is not shown that the given answer is unique. |  |  |
| :---: | :---: | :---: |
| The main idea of the solution is found. The answer is missing or |  |  |
| incorrect. |  |  |