B3. In a red box, there are twelve balls numbered from 1 to 12. Jan moved some of these balls to a green box. He found that for any two balls in the green box, if the numbers on these balls are $a$ and $b$, then the ball with the number $|a-b|$ is in the red box. What is the maximum number of balls Jan could have moved to the green box?

## 55th Mathematical Competition for High School Students in Slovenia

Regional Competition, March 30, 2011

## Solutions for 1st Year

In section A, a correct answer will be awarded two points, while an incorrect answer will result in a deduction of half a point. To avoid a potential negative final score, each contestant is given an initial 2 points.

| A1 | A2 | A3 |
| :---: | :---: | :---: |
| C | C | D |

Explanations:
B3. Jan can move 6 balls. If he chose all the balls with odd numbers, the difference between the numbers on any two balls is an even number that is not written on any of the balls in the green box.

Suppose Jan moved at least 7 balls. Let the numbers on the seven balls in the green box be $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<a_{6}<a_{7}$. Then $a_{7}-a_{1}, a_{6}-a_{1}, a_{5}-a_{1}, a_{4}-a_{1}$, $a_{3}-a_{1}, a_{2}-a_{1}$ are six different natural numbers, each less than 12. The balls with these numbers are in the red box. This is not possible, as there are at most five balls in the red box.

Written that Jan can move 6 balls ...................................................... 2 points Example given with six balls that satisfy the conditions ................. 2 points Start of the treatment of the case where there are at least 7 balls in the green box ............ 1 point (The point is not recognized if it is assumed that there are exactly 7 balls in the green box, and examples with more than seven balls are not provided. The point is recognized if the case with exactly 7 balls is treated and it is justified that this treatment is sufficient.)

Finding that in this case there are six numbers that should be on the balls in the red box .1 point

Conclusion that there cannot be more than 6 balls in the green box 1 point

[^0]:    Writing fractions with a common denominator

    1 point

    Conclusion that $a(a-4)$ is a divisor of $a^{2}+2 a-8$ 1 point

    Conclusion that $a$ (or $a-4$) divides $a^{2}+2 a-8$ 1 point

    Limiting to a finite number of possibilities (for example, $a$ divides 8) 1 point Treating all cases (including negative ones!) 1 point

    Both solutions $a=2$ and $a=-4$ 1 point