## Task B-1.4.

In each cell of a $4 \times 4$ table, a number is written. For each cell, the sum of the numbers in its adjacent cells is equal to the same natural number $x$ (two cells are adjacent if they share a side). Determine the number $x$ such that the sum of all numbers in the table is 282.
## First Solution.

Let's label the fields as follows:

| $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{D}$ |
| :---: | :---: | :---: | :---: |
| $\mathrm{E}$ | $\mathrm{F}$ | $\mathrm{G}$ | $\mathrm{H}$ |
| $\mathrm{I}$ | $\mathrm{J}$ | $\mathrm{K}$ | $\mathrm{L}$ |
| $\mathrm{M}$ | $\mathrm{N}$ | $\mathrm{O}$ | $\mathrm{P}$ |

We will calculate the desired sum by grouping the numbers from the field into disjoint sets.

We will group in sequence for some field in the table its adjacent members and be careful not to count the same number twice:

$$
\begin{array}{cl}
A+B+C+D+E+F+G+H+I+J+K+L+M+N+O+P & \\
=(A+F+C)+(B+D+G)+(E+J+M)+(H+K+P)+(I+N)+(L+O) & 3 \text { points } \\
=x+x+x+x+x+x=6 x & 1 \text { point }
\end{array}
$$

From the condition of the problem, it follows that $6 x=282$, or $x=47$.