2002 Indonesia MO Problems/Problem 5
Problem
Nine of the following ten numbers: are going to be filled into empty spaces in the table shown below. After all spaces are filled, the sum of the numbers on each row will be the same. And so with the sum of the numbers on each column, will also be the same. Determine all possible fillings.
Solution
The sum of the numbers currently in the grid and all ten numbers is . Note that the sum of the numbers in each row and the sum of the numbers in each column are whole numbers, so the sum of all the numbers that would be in the grid is a multiple of and . Because and , we know that a number not in the grid is congruent to modulo and modulo . The only number of the ten numbers that meets the criteria is , so will not be in the grid.
That makes the sum of the numbers in the grid , so the sum of each row is and the sum of each column is . Thus, must appear between and . Also, the remaining two numbers in the middle column sum to , The only two numbers from the remaining list that add up to are and .
If is on the top row of the middle column, then the right column of the top row must have , which couldn't happen. Therefore, is on the top row with to the right of it, and is on the second row of the middle column with to the left of it.
The remaining numbers are , , , and . The remaining numbers of the last row must sum to , so and are in the bottom row while and are on the third row. Also, the remaining numbers of the last column must sum to , so and are on the last column.
With this in mind, there is only one possible configuration that satisfies the conditions (shown below).
See Also
2002 Indonesia MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 6 |
All Indonesia MO Problems and Solutions |