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.

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TopNewestSuppose 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:

In total, there are \(2N + 1 - \lceil \frac{N}{2} \rceil\) colorings. For \(N = 2015\), this gives \(3023\). – Ivan Koswara · 1 year, 3 months ago

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If ball a is green and ball b is green then ball a+b<2015 then A+B IS GREEN – Amrit Nimiyar · 1 year, 3 months ago

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