## Task 2 - 150922

|  | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{D}$ | $\mathrm{E}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{a}$ | 1 | 2 | 3 |  |  |
| $\mathrm{b}$ |  |  |  |  |  |
| $\mathrm{c}$ |  |  |  | 5 |  |
| $\mathrm{d}$ |  |  |  |  | 4 |
| $\mathrm{e}$ |  |  |  |  |  |

The digits 1, 2, 3, 4, and 5 are to be entered into the square such that each digit from 1 to 5 appears exactly once in each row, column, and both diagonals. The already entered digits should not be changed.

a) Provide a corresponding entry that meets the conditions!

b) Investigate whether different entries that meet the conditions are possible, and if so, determine all such entries!

The letters at the edges of the square are intended to facilitate the description of the solution path. For example, the digit 5 is already in field cD, abbreviated as cD:5.
aD:4 (Row a, columns ABC occupied; dE:4)

aE:5 (Row a, columns ABCD occupied)

bB:5 (Main diagonal: aA:1, cC: $\neq 5$ due to cD:5, dd $: \neq 5$ due to cD:5, eE: $\neq 5$ due to aE:5)

eC:5 (Row e, eA: $\neq 5$ due to aE:5, eB: $\neq 5$ due to bB:5, eD: $\neq 5$ due to $\mathrm{cD}: 5$, eE $: \neq 5$ due to aE: 5 )

dA:5 (Row d; in all columns BCDE there are already fives).

cC:4 (Main diagonal: $\mathrm{aA}, \mathrm{bB}$ occupied, $\mathrm{dD}: \neq 4$ due to $\mathrm{aD}: 4$, eE: $\neq 4$ due to $\mathrm{dE}: 4$ )

eB:4 (Column B, rows ab occupied, cB: $\neq 4$ due to cC:4, cd: $\neq 4$ due to dE:4)

bA:4 (Column A, in all other rows there are already fours)

Sketch:

|  | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{D}$ | $\mathrm{E}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{a}$ | 1 | 2 | 3 | 4 | 5 |
| $\mathrm{b}$ | 4 | 5 |  |  |  |
| $\mathrm{c}$ |  |  | 4 | 5 |  |
| $\mathrm{d}$ | 5 |  |  |  | 4 |
| $\mathrm{e}$ |  | 4 | 5 |  |  |

If bD:1, then eA: $\neq 1$ (main diagonal) and eD: $\neq 1$, so eE:1 (since eB and eC are already occupied). This is a contradiction to aA:1 on the main diagonal. Therefore, bD: $\neq 1$.

$\mathrm{eD}: 1$ (Column $\mathrm{D}, \mathrm{bD}: \neq 1, \mathrm{aD}, \mathrm{cD}$ occupied, $\mathrm{dD}: \neq 1$ due to $\mathrm{aA}: 1$ )

$\mathrm{dB}: 1$ (Secondary diagonal, $\mathrm{aE}$, cc occupied, $\mathrm{bD}: \neq 1$ due to $\mathrm{eD}: 1, \mathrm{eA}: \neq 1$ due to aA:1)

cE:1 (Row c, columns CD occupied, columns AB forbidden, as they already have ones)

bC:1 (Row b, in all other columns there are already ones)

Sketch:

|  | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{D}$ | $\mathrm{E}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{a}$ | 1 | 2 | 3 | 4 | 5 |
| $\mathrm{b}$ | 4 | 5 | 1 |  |  |
| $\mathrm{c}$ |  |  | 4 | 5 | 1 |
| $\mathrm{d}$ | 5 | 1 |  |  | 4 |
| $\mathrm{e}$ |  | 4 | 5 | 1 |  |

dC:2 (Column C, rows abce occupied) $\quad ; \quad$ dD:3 (Row d, columns ABCE occupied) bD:2 (Column D, rows acde occupied) ; bE:3 (Row b, columns ABCD occupied) eE:2 (Column E, rows abcd occupied) ; eA:3 (Row e, columns BCDE occupied) cA:2 (Column A, rows abde occupied) ; cB:3 (Row c, columns ACDE occupied)

Thus, the following table is necessarily derived:

|  | $\mathrm{A}$ | $\mathrm{B}$ | $\mathrm{C}$ | $\mathrm{D}$ | $\mathrm{E}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{a}$ | 1 | 2 | 3 | 4 | 5 |
| $\mathrm{b}$ | 4 | 5 | 1 | 2 | 3 |
| $\mathrm{c}$ | 2 | 3 | 4 | 5 | 1 |
| $\mathrm{d}$ | 5 | 1 | 2 | 3 | 4 |
| $\mathrm{e}$ | 3 | 4 | 5 | 1 | 2 |

That this also meets all the conditions of the problem statement is shown by the verification. Therefore, there is only this one variant to fill the table according to the specifications.