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PrintTHE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD
Romania counting and probability
Problem
Determine the smallest positive integer such that, for any coloring of the elements of the set with two colors, the equation has a monochrome solution with . (We say that the equation has a monochrome solution if there exist distinct, having the same color, such that .)
Solution
For there exists a coloring of the numbers from with two colors such that the equation has no monochrome solution: we color the numbers from with one color, and the elements of with the second color. For we color with the first color the numbers from and with the second one the elements of . Again, the equation has no monochrome solution. It follows that . We prove that, for any coloring with two colors of the elements of the set , the equation has monochrome solutions. Assume the contrary: there exists a coloring such that there is no monochrome solution to the equation. Case 1: If , , have color 1, then , , need to have color 2, hence , , need to be of color 1. But then , and the equation has a monochrome solution (of color 1).
Case 2: If and are of color 1 while has color 2, then has color 2, and has color 1. But , and numbers and have color 1, hence needs to be of color 2. Similarly, as and , have color 1, it follows that has color 2. If has color 1, then leads to a monochrome solution. If has color 2, then is a monochrome solution.
Case 3: If and have color 1 while has color 2, then has color 2, has color 1. As and have color 1, we need to have color 2. Similarly, and are of color 1, therefore has color 2, has color 1. But , hence we have a monochrome solution (of color 1 or color 2, depending on the color of ).
Case 4: has color 1, and have color 2. Then has color 1. But and having color 1 means that needs to be of color 2. This leads to and, again, regardless on the color of , the equation has a monochrome solution.
Case 2: If and are of color 1 while has color 2, then has color 2, and has color 1. But , and numbers and have color 1, hence needs to be of color 2. Similarly, as and , have color 1, it follows that has color 2. If has color 1, then leads to a monochrome solution. If has color 2, then is a monochrome solution.
Case 3: If and have color 1 while has color 2, then has color 2, has color 1. As and have color 1, we need to have color 2. Similarly, and are of color 1, therefore has color 2, has color 1. But , hence we have a monochrome solution (of color 1 or color 2, depending on the color of ).
Case 4: has color 1, and have color 2. Then has color 1. But and having color 1 means that needs to be of color 2. This leads to and, again, regardless on the color of , the equation has a monochrome solution.
Final answer
13
Techniques
Coloring schemes, extremal arguments