Michel must write the numbers from 1 to 7 in each of the circles in the figure below, so that the results obtained by adding the 3 numbers at the vertices of triangles $X, Y$, and $Z$ are 3 consecutive numbers. Which number can be in the central circle? Find all possibilities.

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-36.jpg?height=536&width=595&top_left_y=1683&top_left_x=612)
Solution

Let's denote the numbers that will be written in the circles by the letters $a, b, c, d, e, f$ and $g$, as shown in the figure below.

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-37.jpg?height=531&width=595&top_left_y=549&top_left_x=802)

Suppose the three consecutive numbers obtained from the sums of the vertices of triangles $X, Y$ and $Z$ are $k, k+1$ and $k+2$. Thus,

$$
\begin{aligned}
3 k+3 & =k+(k+1)+(k+2) \\
& =(a+c+d)+(b+d+e)+(d+f+g) \\
& =(a+b+c+d+e+f+g)+2 d
\end{aligned}
$$

Since in the 7 circles we must find each number from the set $\{1,2,3,4,5,6,7\}$, then

$$
a+b+c+d+e+f+g=1+2+3+4+5+6+7=28
$$

Therefore,

$$
3 k+3=28+2 d
$$

Since the left-hand side of this equation is an integer multiple of 3, the same must be true for the right-hand side. Substituting the possible values of $d$, we have the following options for $28+2 d$:

$$
2 \cdot 1+28,2 \cdot 2+28,2 \cdot 3+28,2 \cdot 4+28,2 \cdot 5+28,2 \cdot 6+28,2 \cdot 7+28
$$

In this list, only $2 \cdot 1+28,2 \cdot 4+28$ and $2 \cdot 7+28$ are multiples of 3, so in principle, $d$ can only be 1, 4, or 7 (note that all leave a remainder of 1 when divided by 3). To show that these values of $d$ are admissible, for each of them, we need to exhibit an example of a configuration satisfying the statement. From the last equation, we can conclude that

$$
\begin{aligned}
& d=1 \quad \rightarrow \quad k=9 \\
& d=4 \quad \rightarrow \quad k=11 \\
& d=7 \quad \rightarrow \quad k=13
\end{aligned}
$$