5. Andrey thought of 4 integers and told Vanya all their pairwise sums. Vanya remembered only 4 of them accurately: $0, 3, 6$ and 18. About the two remaining sums, Vanya only remembered that they were two-digit natural numbers. What numbers did Andrey think of?
Answer: $-6 ; 6 ; 9 ; 12$.

Solution. Let the numbers (in ascending order) be a; b; c; d. Considering that the two remaining sums are two-digit, the minimum sums are 0, 3, and 6. Therefore, $\mathrm{a}+\mathrm{b}=0 ; \mathrm{a}+\mathrm{c}=3$, and either $\mathrm{a}+\mathrm{d}=6$ or $\mathrm{b}+\mathrm{c}=6$. From the first two equations, we find $\mathrm{b}=-\mathrm{a} ; \mathrm{c}=3-\mathrm{a}$. We get $\mathrm{b}+\mathrm{c}=-\mathrm{a}+3-\mathrm{a}=3-2 \mathrm{a}$. Since $\mathrm{a}$ is an integer, $\mathrm{b}+\mathrm{c}$ is odd, so it cannot be equal to 6, which means $\mathrm{a}+\mathrm{d}=6$ and $\mathrm{d}=6-\mathrm{a}$.

Of the three remaining sums (b+c, b+d, c+d), one must be equal to 18.

$\mathrm{b}+\mathrm{c}=3-2 \mathrm{a}$ - odd

$b+d=-a+6-a=6-2 a$ - even

$\mathrm{c}+\mathrm{d}=3-\mathrm{a}+6-\mathrm{a}=9-2 \mathrm{a}$ - odd

Of the three sums, only one is even, so $b+d=6-2 a=18$, from which $a=-6$. It is easy to verify by checking that the two remaining sums are indeed two-digit.

Criteria. Answer without explanation -1 point. Equations $\mathrm{a}+\mathrm{b}=0$; $\mathrm{a}+\mathrm{c}=3$ are set up, but no further progress -2 points.