5. On the board, five integers are written. By adding them in pairs, the following sums are obtained: $0 ; 2 ; 4 ; 4 ; 6 ; 8 ; 9 ; 11 ; 13 ; 15$. Determine the integers!

Can it be, starting from five integers and adding them in pairs, that the following sums are obtained: 12;13;14;15;16;16;17;17;18;20?
Solution. Let's note that among the five integers, there are no two equal numbers, because otherwise, we would have more than two pairs of equal sums. Therefore, the required integers are distinct from each other.

If we sum the first given sums, we get

$$
0+2+4+4+6+8+9+11+13+15=72
$$

From the condition of the problem, each of the five integers written on the board appears exactly four times in this sum, so the sum of the five integers is equal to $72: 4=18$.

Let's denote the five integers by $a, b, c, d, e$ and assume

$$
a<b<c<d<e
$$

From the given sums, we have that the smallest sum is 0, the largest sum is 15, and the sum of the five integers is 18, so it must be

$$
c=18-(a+b)-(d+e)=18-0-15=3 \text{ i.e. } a<b<3<d<e
$$

According to the values of the given sums, the second sum in order is obtained from the first and third integers, i.e., $a+c=2$, so $a=2-3=-1$. Since the sum of the first two integers is equal to 0, it follows that the second integer is equal to 1, i.e., $b=0-(-1)=1$. Similarly, the second-to-last sum in order is obtained from the third and last integers, i.e., $c+e=13$, so $e=13-3=10$. Finally, from the last sum, we get $d=15-10=5$. Therefore, the required numbers are: $-1 ; 1 ; 3 ; 5 ; 10$.

If, however, we sum the second given sums, we get

$$
12+13+14+15+16+16+17+17+18+20=158
$$

But the number 158 is not divisible by 4, so such integers do not exist.

## VIII Department