If I'm not wrong,there must be some problem which is similar as this. This appears in another competition. (It only changes the number of it) But I forget where.

Dear mathlinkers, at the site http://www.oei.es/oim/ibero2009.pdf you can find the solutions (in spanish) of the problems proposed past september in the contest

gb2124 wrote: If I'm not wrong,there must be some problem which is similar as this. This appears in another competition. (It only changes the number of it) But I forget where. That's correct, This problem can be generalized. I obtain that the answer is 89, and my answer was correct!

The answer is 89. Firstly, we will make some assumptions and conventions. A $ L$-arc is a set of $ L$ consecutive points. The orientation is clockwise. Suppose that, for some coloring, no 89-arc has all 10 colors. So we will prove these facts: 1 - For each 11-arc there is one color that doesn't appear among the 89 following points; 2 - Each 12-arc has just 2 colors. If 1) is not valid for some 11-arc, then the 89 next points has all 10 colors (all 100-arc has all colors), contradiction! For 2): split any 100-arc in eight 11-arcs ($ A_1,A_2, \ldots,A_8$) and a 12-arc $ A_9$. Using 1), we know that for the arcs $ A_i, i=1,2,3,4,5,6,7,8$ there is a color $ i$ that doesn't appear at $ A_{i+1},\ldots,A_9$. Clearly all colors $ 1,2,\ldots,8$ are distinct, so there is just 2 colors at $ A_9$. This argument works for any 12-arc. So, that there is at most 2 color at any 12-arc. Now we need to prove that the "at most" is, in fact, "exactly". If we get a monochromatic 12-arc, just see the 88-arc following it. The 100-arc obtained "glue-ing" them has 10 colors, so the last point of the 12-arc and all 88-arc form a 89-arc with all 10 colors, contradiction! Let it $ X_0$ a point with color 0, this color not appear at the 89-arc following it. The 11-arc following $ X_0$ form the arc $ A_1$. The arc $ X_0+A_1$ is a 12-arc, with 2 colors, namely 0 (the color of $ X_0$) and 1. By the way, $ X_0$ is the only point at this arc with the color 0. So the 11-arc $ A_1$ has color 1. If $ X_1$ is the latest point of $ A_1$, the 89-arc following doesn´t have the 1 color at it. Applying the same reasoning above, we can see a beautiful pattern: monochromatic 11-arcs alternating this colors. But hey!, 6000 is not a multiple of 11. It is impossible! So, we always obtain a 89-arc with all 10 colors. The problem now is to exhibit a colouring such that a 88-arc does not have the property. I will do it later(sloth...)