
2. The numbers 1 to 25 are each coloured blue or red. Determine all possible colourings that satisfy the following rules:

- The number 5 is red.
- If numbers $x$ and $y$ have different colours and $x+y \leqslant 25$, then $x+y$ is blue.
- If numbers $x$ and $y$ have different colours and $x \cdot y \leqslant 25$, then $x \cdot y$ is red.

## 2. Version for klas $5 \&$ klas 4 and below

We first consider the case that 1 is red. Then all numbers from 1 to 25 are red. Indeed, suppose that some number $k$ is blue. Then 1 and $k$ have different colours, hence by the third rule the number $1 \cdot k=k$ must be red. But this contradicts the assumption that it was blue. Observe that colouring all numbers red indeed gives a correct colouring.

Now consider the case that 1 is blue. Since 5 is not blue, it follows from the second rule that 1 and 4 have the same colour. Hence, 4 is blue. Similarly, 2 and 3 must have the same colour. In fact, both must be blue since if 2 were red, then $3=1+2$ would have to be blue because of the second rule.

We now know that 1, 2, 3, and 4 are blue. By the second rule, we see that also $6=5+1$, $7=5+2,8=5+3$, and $9=5+4$ are blue. Applying the second rule again, we find in turn that $11,12,13,14$ are blue, that $16,17,18,19$ are blue, and that $21,22,23,24$ are blue.

Because of the third rule, $10=2 \cdot 5,15=3 \cdot 5$, and $20=4 \cdot 5$ are red. Only the colour of 25 is not yet determined by the rules. We find two possible colourings:

(1) Only $5,10,15,20,25$ (the numbers divisible by 5 ) are red.

(2) Only 5, 10,15, 20 are red.

We check that colouring (1) is indeed correct. The second rule is satisfied, because the sum of a number divisible by 5 and a number not divisible by 5 is itself not divisible by 5 . The third rule is satisfied since the product of a number divisible by 5 and a number not divisible by 5 is itself divisible by 5 .

Colouring (2) differs from colouring (1) only in the colour of 25 . To see that colouring (2) is correct, we only need to check the rules in the situation that one of the numbers involved is 25 . We check that 25 is not the product of a blue and a red number (that is correct). We must also check that there is no red number $k$ such that $25+k$ is red or $25 k$ is blue. That is automatically true since the numbers greater than 25 are not coloured.

We conclude that there are 3 correct colourings altogether: the colouring in which all numbers are red, colouring (1), and colouring (2).

