The eight points A, B,.., G and $H$ lie on five circles as shown in the figure. Each of these letters is replaced by one of the eight numbers 1, 2, ..., 7, and 8 such that the following conditions are met:

(i) Each of the eight numbers is used exactly once.

(ii) The sum of the numbers on each of the five circles is the same.

![](https://cdn.mathpix.com/cropped/2024_04_17_b65d8712533406496cc8g-3.jpg?height=705&width=713&top_left_y=204&top_left_x=680)

(Walther Janous)

Answer. There are 8 possibilities.
Let $a, b, \ldots, g$ and $h$ be the numbers that replace the points $A, B, \ldots, G$ and $H$. Since the sums on all circles must equal $a+b+c+d$, it immediately follows that $e=c+d$, $f=d+a$, $g=a+b$, and $h=b+c$. This gives us $e+f+g+h=2(a+b+c+d)$. On the other hand,

$$
36=1+2+3+4+5+6+7+8=(a+b+c+d)+(e+f+g+h)=3(a+b+c+d)
$$

thus $a+b+c+d=12$. For this sum, only the values 1,2,3,6 or 1, 2, 4, 5 are possible.

If we replace the point $A$ with the number 1, then $b \neq 2$ must hold, because otherwise, due to $a+b+e=1+2+e=12$, the value $e=9$ would be inadmissible. Similarly, $d \neq 2$ must hold. Therefore, $c=2$. Then the cases for $b$ and $d$ are $b=3$ and $d=6$, or $b=6$ and $d=3$, or $b=4$ and $d=5$, or $b=5$ and $d=4$. In the last two cases, however, $f=h=6$ or $e=g=6$ would result, which is not admissible. In the first two cases, the remaining numbers are uniquely determinable from the sum of the numbers in the corresponding circles.

Thus, when replacing $A$ with 1, we get exactly two possibilities for replacing the remaining points. Similarly, we could also replace the points $B, C$, or $D$ with 1, resulting in a total of $4 \cdot 2=8$ possibilities.

(Reinhard Razen)