
2. The numbers 1 to 15 are each coloured blue or red. Determine all possible colourings that satisfy the following rules:

- The number 15 is red.
- If numbers $x$ and $y$ have different colours and $x+y \leqslant 15$, then $x+y$ is blue.
- If numbers $x$ and $y$ have different colours and $x \cdot y \leqslant 15$, then $x \cdot y$ is red.

## 2. Version for klas 6

We first consider the case that 1 is red. Then all numbers from 1 to 15 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 we consider the case that 1 is blue. Observe that when two numbers sum to 15, those numbers must have the same colour by the second rule. We get the following pairs of numbers of the same colour: 1 and 14, 2 and 13, 3 and 12,4 and 11,5 and 10,6 and 9, 7 and 8 .

The number 2 is blue. Indeed, suppose that 2 is red. From the second rule, we derive that $3=1+2$ is blue. Repeatedly applying the same rule we find that $5=3+2$ is blue, that $7=5+2$ is blue, and finally that 15 is blue. Since 15 is in fact not blue, 2 cannot be red.

The number 7 is blue. Indeed, suppose that 7 is red. It then follows from the second rule that $8=1+7$ is blue. However, 7 and 8 must have the same colour, so this cannot be the case.

The number 4 is blue. Indeed, suppose that 4 is red. It then follows from the second rule that $11=4+7$ is blue. But 4 and 11 have the same colour, so this cannot be the case.

Recall, that 3 and 12 have the same colour, and so do 6 and 9. In fact, all four numbers must have the same colour. Indeed, otherwise $9=3+6$ would be blue by the second rule and also $12=3+9$ would be blue by the second rule.

So far, we know that 15 is red, that $1,2,4,7,8,11,13,14$ are blue, that $3,6,9,12$ have the same colour, and that 5,10 have the same colour. The numbers 3 and 5 cannot both be red, since if 3 is red, $5=2+3$ is blue by the second rule. The three remaining colour combinations for the numbers 3 and 5 result in the following three colourings:

(1) Only 15 is red.

(2) Only $5,10,15$ (the numbers divisible by 5 ) are red.

(3) Only $3,6,9,12,15$ (the numbers divisible by 3 ) are red.

It is easy to check that all three colourings are indeed correct. We write this out for the third colouring, the other two can be checked in a similar way. That the sum of a red and a blue number is always blue follows from the fact that the sum of a number divisible by 3 and a number not divisible by 3 is itself not divisible by 3 . That the product of a blue and a red number is always red follows from the fact that the product of a number divisible by 3 and a number not divisible by 3 is itself divisible by 3 .

We conclude that there are 4 correct colourings in total: the colouring in which all numbers are red, and colourings (1), (2), and (3).
