XXVII Olimpiada Matemática Rioplatense

In order for the number in the first column to be a multiple of $5$, we must write $5$ in the cell corresponding to its units, that is, in row $3$ and column $1$. The digits in the cells corresponding to the units of the numbers that are multiple of $2$, $4$ and $6$ must be even. Then, the number in the third row (which is a multiple of $4$) begins with $5$ and its remaining two digits are even. The multiples of $4$ satisfying this condition, with no repeated digits and without $0$, are $$524,528,548,564,568,584.$$ In particular, we deduce that the digit in row $3$, column $3$ can only be $4$ or $8$.

		even
5	even	4 - 8

For the number in the third column, we look for the multiples of $7$ beginning with an even digit, ending with $4$ or $8$, with no repeated digits, and without $0$ and $5$ among its digits. There are two: $238$ and $294$. Combining them with the numbers listed above, that are the candidates for the third row (note that $524$ and $528$ cannot be used, since $2$ must be written in the first row), we obtain the following possibilities:

	2
	3
5	4	8

Board 1

	2
	3
5	6	8

Board 2

	2
	9
5	6	4

Board 3

	2
	9
5	8	4

Board 4

To determine the values of the remaining digits, we take into account that the sums of the digits in row $2$ and the sum of the digits in column $2$ must be multiples of $3$ (in order that the corresponding numbers are multiple of $3$ and $6$, respectively).

Board 1: The remaining digits are $1$, $6$, $7$, $9$; thus, two are congruent with $1$ modulo $3$ and two are congruent with $0$ modulo $3$. We write the board modulo $3$:

a	b	2
c	d	0
2	1	2

If $d=1$, the number in the second row cannot be a multiple of $3$ and, if $d=0$, the number in the second column cannot be a multiple of $3$. Then, there is no solution for Board 1.

Board 2: The remaining digits are $1$, $4$, $7$, $9$; one is congruent with $0$ modulo $3$ and three are congruent with $1$ modulo $3$. As before, by writing the board modulo $3$

a	b	2
c	d	0
2	0	2

and considering both possible values of $d$, it follows that it is not possible to fill in the board satisfying the required conditions.

Board 3: The remaining digits are $1$, $3$, $7$, $8$; two are congruent with $1$ modulo $3$, one is congruent with $0$ modulo $3$ and the remaining one is congruent with $2$ modulo $3$. There is a unique way to fill in the board modulo $3$:

0	1	2
1	2	0
2	0	1

which leads to the following two solutions:

3	1	2
7	8	9
5	6	4

3	7	2
1	8	9
5	6	4

Board 4: The remaining digits are $1$, $3$, $6$, $7$; two are congruent with $0$ modulo $3$, and two are congruent with $1$ modulo $3$. There is a unique way to fill in the board modulo $3$:

1	1	2
0	0	0
2	2	1

leading to the following solutions:

1	7	2
3	6	9
5	8	4

7	1	2
3	6	9
5	8	4

1	7	2
6	3	9
5	8	4

7	1	2
6	3	9
5	8	4