Problem 4.6. Fill in the letters $A, B, C, D, E$ with the digits $1,2,3,4,5$ so that the sum of the digits in all $1 \times 3$ rectangles (both horizontal and vertical) equals 13. Each of the digits from 1 to 5 should appear in the table exactly once.

| $\mathbf{7}$ | $A$ |  |
| :---: | :---: | :---: |
| $B$ | $\mathbf{6}$ | $C$ |
| $D$ | $E$ | $\mathbf{8}$ |

(a) Instead of the letter $A$
(1) the digit 1 should be placed
(b) Instead of the letter $B$
(2) the digit 2 should be placed
(c) Instead of the letter $C$
(3) the digit 3 should be placed
(d) Instead of the letter $D$
(4) the digit 4 should be placed
(e) Instead of the letter $E$
(5) the digit 5 should be placed
Answer: a3 b5 c2 d1 e4.

Solution. The sum of all numbers in the table is $1+2+3+4+5+6+7+8=36$, and in each of the first two columns it is 13. Therefore, the total sum of the numbers in the table is $36=13+13+(8+C)$, from which $C=2$.

In the second row, the sum is $13=B+6+C=B+6+2$, from which $B=5$.

In the first column, the sum is $13=7+B+D=7+5+D$, from which $D=1$.

In the bottom row, the sum is $13=D+E+8=1+E+8$, from which $E=4$.

In the second column, the sum is $13=A+6+E=A+6+4$, from which $A=3$.

Thus, the required arrangement is as follows:

| 7 | 3 |  |
| :--- | :--- | :--- |
| 5 | 6 | 2 |
| 1 | 4 | 8 |
|  |  |  |