Two marksmen made 5 shots each, and the hits were as follows: $10,9,9,8,8,5,4,4,3,2$. With their first three shots, they scored the same number of points, but with their last three shots, the first marksman scored three times more points than the second.

How many points did each of them score with their third shot?
Let $a_{i}$ denote the number of points scored by the first shooter on the $i$-th shot, and $b_{i}$ the number of points scored by the second shooter on the $i$-th shot. Then

$a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}, a_{3}+a_{4}+a_{5}=3\left(b_{3}+b_{4}+b_{5}\right)$

By the condition $b_{3}+b_{4}+b_{5} \geq 2+3+4=9$, and the sum $a_{3}+a_{4}+a_{5} \leq 10+9+9=28$ and is divisible by 3. Therefore, $\left\{b_{3}, b_{4}, b_{5}\right)=$ $\{2,3,4\}$ $\left\{a_{3}, a_{4}, a_{5}\right\}=\{10,9,8\}$, and $\left\{a_{1}, a_{2}, b_{1}, b_{2}\right\}=\{4,5,8,9\}$. If among the numbers $a_{1}, a_{2}$ there is an 8 or 9, then $a_{1}+a_{2}+a_{3} \geq 4$ $+8+8=20$, while $b_{1}+b_{2}+b_{3} \leq 9+5+4=18$. Therefore, $\left\{a_{1}, a_{2}\right\}=\{4,5\}$, and $\left\{b_{1}, b_{2}\right\}=\{8,9\}$. Consequently, $a_{3}-$ $b_{3}=8+9-(4+5)=8$, that is, $a_{3}=10, b_{3}=2$.

## Answer

The first shooter scored 10 points, and the second shooter scored 2.