Jancsi and Feri each fired 5 shots at the same 12-ring target during a small shooting competition. On the target, we see 1 hit in the 5th, 8th, 11th, and 12th rings, and 2 hits in the 7th, 9th, and 10th rings. We also heard during the shots that each of them had one 10-ring hit. We learned that Jancsi scored 7 times as many points (rings) with his last four shots as with his first; Feri scored 5 times as many points with his first four shots as with his last, and his last two shots scored as many rings as his first two. - Whose was the 12-ring hit? What could have been the scores of the two boys, and the order of Feri's hits?
I. solution. The value of the 10 shots in descending order is $12,11,10,10,9,9,8,7,7,5$ points, their sum is 88. Jancsi's last 4 shots could score at most $12+11+10+9=42$ points (since the other 10 is by Feri), so his first shot was not more than $42: 7=6$ points. Thus, his first shot was a hit in the 5-point ring, the sum of his other shots is 35 points - his total score is 40 points - and the sum of his 3 unknown shots, excluding the 10-point ring, is 25 points. At least one of these is odd, and it can only be a 7, otherwise, he would have at least $8+9+9=26$ points from them. The remaining 18 points could have been scored in two shots in two ways: $7+11$ or $9+9$, so Jancsi's score is either $5,10,7,7,11$ or $5,10,7,9,9$, and the 12-point ring was definitely hit by Feri.

48 points remained for Feri, so his last shot is one-sixth of the sum, an 8. He also had the 10 and the 12, and the remaining two shots were either $9+9$ or $7+11$. The point value of his middle shot is one of the even numbers, because it is derived by subtracting twice the point value of the last two shots from 48. By choosing the point value of the middle shot and subtracting 8 from the remaining half, the point value of the 4th shot is derived, and if such a hit was among the others, the remaining two shots fit into the first two positions. If these are different, they can follow each other in two orders. 5 possible orders arise:

$9,9,12,10,8 ; \quad 7,12,10,11,8 ; 12,7,10,11,8 ; 7,11,12,10,8 ; 11,7,12,10,8$

Lintner Ferenc (Budapest, Apáczai Csere J. Gyak. G. I. o. t.)

II. solution. Let the value of Jancsi's first shot be $j_{1}$ points, and Feri's 5th shot be $f_{5}$ points, so their scores are $(1+7) j_{1}$ and $(5+1) f_{5}$, respectively, and

$$
8 j_{1}+6 f_{5}=88, \quad \text { from which } \quad j_{1}=11-\frac{3 f_{5}}{4} \text {. }
$$

Accordingly, $f_{5}$ is divisible by 4, so it is either 8 or 12. The latter would yield $j_{1}=2$, but there is no such shot, so $f_{5}=8$, and $j_{1}=5$, Feri scored 48, and Jancsi 40 in total.

The combined value of Feri's first two and last two shots $\left(f_{1}+f_{2}\right)+\left(f_{4}+f_{5}\right)$ is twice $\left(f_{4}+f_{5}\right)$, so it is even, thus his 3rd shot, $f_{3}$, is also of even value. It could have been a 10 or a 12, because then the value of the extreme shot pairs is 19 or 18 points, so $f_{4}=11$, or 10 points, and each case can be continued from the given shots to the complete sequence of hits:

$$
f_{3}=10 \quad \text { in case } \quad f_{4}=11, \quad f_{1}+f_{2}=7+12=12+7
$$

(it cannot be $f_{1}+f_{2}=10+9$, because the second 10 is by Jancsi);

$$
f_{3}=12 \text { in case } f_{4}=10, f_{1}+f_{2}=7+11=11+7 \text { or } 9+9 \text {. }
$$

Accordingly, the 12-point hit is definitely by Feri. According to the last possibility, the order of Feri's shots is uniquely $9,9,12,10,8$, and each of the previous two possibilities could have arisen in two orders. Jancsi's score in the case of Feri's first two possibilities is $5,7,9,9,10$, in the last case 5, 7, 7, 10, 11 (only the order of the 5-point hit is certain). 

Süttő Klára (Budapest, Ságvári E. Gyak. G. I. o. t.)