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# That's a whole bunch of balls

A collection of 2015 balls are numbered from 1 to 2015. Each ball may be coloured green or red according to the following rules:

(i) If balls $$a$$ and $$b$$ are two different red balls with $$a+b \leq 2015$$, the ball $$a+b$$ is also red.

(ii) If ball $$a$$ is red and another ball $$b$$ is green with $$a+b \leq 2015$$, then $$a+b$$ is green.

Find, with proof, the number of different possible ways of colouring the 2015 balls.

Bonus: Generalise this for $$n$$ balls.

Note by Sharky Kesa
1 year, 7 months ago

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Suppose there are $$N$$ balls.

All green is one possible coloring. Otherwise, there exists a red ball; let $$m$$ be the smallest red ball. We claim that all balls with number not a multiple of $$m$$ are green. Suppose there exists another red ball with number not divisible by $$m$$; let the smallest of such numbers be $$n$$. Since $$m$$ is the smallest red ball, $$n > m$$; also, since $$n$$ is not divisible by $$m$$, $$n-m \neq m$$. Thus $$n-m$$ is green. But $$m$$ is red, so $$n = (n-m)+m$$ should be green, contradiction.

A single red ball is also a possible coloring. Otherwise, there exists a second red ball. As shown above, this red ball must be a multiple of $$m$$; call the smallest one to be $$km$$. If $$k \ge 3$$, we can use the same argument as above to reach a contradiction, so $$k = 2$$. Now we claim that all multiples of $$m$$ are red. Suppose there exists a green ball that is a multiple of $$m$$; call the smallest one to be $$km$$. Since $$m, 2m$$ are red by assumption above, $$k \ge 3$$, so $$k-1 \ge 2$$. So $$m, (k-1)m$$ are different balls, and since $$km$$ is the smallest green ball, $$(k-1)m$$ is red. But then $$km = m + (k-1)m$$ should be green, contradiction.

We have determined all the possible colorings above:

• All green: $$1$$ coloring
• One red ball: $$N$$ colorings
• Many red balls: $$N$$ colorings, minus $$\lceil N/2 \rceil$$ where there is only one red ball (only one multiple)

In total, there are $$2N + 1 - \lceil \frac{N}{2} \rceil$$ colorings. For $$N = 2015$$, this gives $$3023$$. · 1 year, 7 months ago