Exercise 10. In each cell of a square table, a strictly positive integer is written. It is assumed that, for each row of the table, the sum of the integers written in the cells of that row is an odd integer. Show that there exists a column of the table such that the sum of the integers in the cells of that column is also odd.
Solution
Solution to Exercise 10 If the sum of the terms in each row is odd and there are 5 rows, the sum of the integers in the 25 cells of the table is itself odd (as the sum of an odd number of odd numbers).
If the sum of the integers in each column is even, then the sum of the 25 cells of the table is itself even. This leads to a contradiction, which means that the sum of the cells in at least one column is odd.
Alternative Solution If the sum of the integers in each row is an odd integer, this means that each row contains an odd number of odd integers.
Since there are five rows, this implies that there is a total of an odd number of odd integers in the grid.
Therefore, there is a column in the grid that contains an odd number of odd integers. Indeed, if each column contained an even number of odd integers, there would be an even number of odd integers in total in the grid.
The sum of the integers in this column is then odd, as required.
Grader's Comment The exercise was generally well done, most students saw the main ideas of the problem. However, the writing is often too imprecise, important arguments for solving the problem are left unsaid, which often cost points. It is also important to remember that there is no reason for the sums of the rows/columns to be all equal, so arguments that say the sum of five odd numbers is odd because 5 times an odd number is odd are not valid. More generally, it is better to do more than less, and to recall the basic arguments of the proof is a good reflex.
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