[ Case Analysis ]  [ Proof by Contradiction ]

In the cells of a $3 \times 3$ table, numbers are arranged such that the sum of the numbers in each column and each row is zero. What is the smallest number of non-zero numbers that can be in this table, given that this number is odd? 

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## Example.

| 0 | -1 | 1 |
| :---: | :---: | :---: |
| -1 | 2 | -1 |
| 1 | -1 | 0 |

Evaluation. We will prove that it is impossible to use fewer non-zero numbers.

If the table contains exactly one non-zero number, then the sum of the numbers in the row containing this number is not zero. Suppose the table contains exactly three non-zero numbers. If all of them are in the same row, then the sum of the numbers in any column is not zero. If not all of them are in the same row, then in some row there is exactly one non-zero number, and the sum of the numbers in this row is not zero.

Suppose the table contains exactly five non-zero numbers. Then there are four zeros in the table, which means that some two zeros are in the same row. Since the sum of the numbers in this row is zero, all the numbers in this row are zeros.

It remains to note that in the column where the remaining zero is located, there are exactly two zeros, which is impossible.

## Answer

7 numbers.

A five-digit number is called irreducible if it cannot be factored into the product of two three-digit numbers.

What is the maximum number of consecutive irreducible five-digit numbers?

## Solution

Note that all numbers divisible by 100 are reducible. Therefore, more than 99 irreducible numbers in a row is impossible. On the other hand, between the numbers $100 \cdot 100$ and $100 \cdot 101$, all numbers are reducible, and there are exactly 99 of them.

## Answer

99.