# Problem 5. (3 points)

In the cells of a $7 \times 7$ table, pairwise distinct non-negative integers are written. It turns out that for any two numbers in the same row or column, the integer parts of their quotients when divided by 8 are different. What is the smallest value that the largest number in the table can take?
# Answer: 54

Solution: Each row must contain seven numbers that give different incomplete quotients when divided by 8. This means that in each row, there is at least one number with an incomplete quotient of 6 or more when divided by 8 (6, not 7, since the minimum value of the incomplete quotient is 0). Therefore, the table contains at least 7 numbers that are not less than $6 \cdot 8=48$. The largest of these is not less than $48+6=54$.

This is indeed possible if numbers with the same incomplete quotients are arranged along the diagonals:

| $0 \cdot 8+1$ | $1 \cdot 8+1$ | $2 \cdot 8+1$ | $3 \cdot 8+1$ | $4 \cdot 8+1$ | $5 \cdot 8+1$ | $6 \cdot 8+1$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $6 \cdot 8+2$ | $0 \cdot 8+2$ | $1 \cdot 8+2$ | $2 \cdot 8+2$ | $3 \cdot 8+2$ | $4 \cdot 8+2$ | $5 \cdot 8+2$ |
| $5 \cdot 8+3$ | $6 \cdot 8+3$ | $0 \cdot 8+3$ | $1 \cdot 8+3$ | $2 \cdot 8+3$ | $3 \cdot 8+3$ | $4 \cdot 8+3$ |
| $4 \cdot 8+4$ | $5 \cdot 8+4$ | $6 \cdot 8+4$ | $0 \cdot 8+4$ | $1 \cdot 8+4$ | $2 \cdot 8+4$ | $3 \cdot 8+4$ |
| $3 \cdot 8+5$ | $4 \cdot 8+5$ | $5 \cdot 8+5$ | $6 \cdot 8+5$ | $0 \cdot 8+5$ | $1 \cdot 8+5$ | $2 \cdot 8+5$ |
| $2 \cdot 8+6$ | $3 \cdot 8+6$ | $4 \cdot 8+6$ | $5 \cdot 8+6$ | $6 \cdot 8+6=54$ | $0 \cdot 8+6$ | $1 \cdot 8+6$ |
| $1 \cdot 8+0$ | $2 \cdot 8+0$ | $3 \cdot 8+0$ | $4 \cdot 8+0$ | $5 \cdot 8+0$ | $6 \cdot 8+0$ | $0 \cdot 8+7$ |

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