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A big equilateral triangle is made up of smaller equilateral triangles. The relation is for n order of the bigger triangle, the number of inner triangles are n^2. Example of such a triangle with n = 3.

enter image description here Now for the given triangles, we mark every vertex, except one, with value of 1, and the remaining one being -1. This one anomaly vertex could be any vertex. In our given example of n being 3, there are 10 vertices, A being one of them. Now at each step, we select one small triangle and reverse the values of all its vertices i.e. 1 becomes -1 and vice versa.

Is it possible to transform all vertex to have value 1?

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This is not possible. The crux of the solution is to partition the vertices into sets so that each small triangle contains a vertex in each set. We will partition the vertices by labeling them as follows (line them up with the horizontal lines in the figure).

1

2 3

3 1 2

1 2 3 1 (this figure stops on this row)

2 3 1 2 3

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.

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Let $S_i$ be the sum of the values assigned to vertices not labeled with $i$. This is invariant modulo 4 since there are exactly two in each small triangle. The vertex originally labeled with -1 will belong to exactly two of these sets, and neither will be able to achieve all ones by the invariant noted above.

Some intuition: vertices numbered the same are on opposite sides of an edge shared by two small triangles on which they lie. By performing the switching operation on both triangles, the aforementioned vertices will switch sign, while all others will remain invariant.

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