1-33

We colored each of the numbers from 1 , 2 , 3 , , 32 , 33 1, 2, 3, \dots, 32, 33 We used 11 11 colors, and each color was used for exactly 3 3 numbers. Is it possible that for the numbers with the same color ( a < b < c a<b<c ) a + b = c a+b=c is always true?

No, it's not possible. Yes, it's possible.

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1 solution

Suppose it is possible. Then if x , y x, y and z z are colored the same, then: x + y = z x+y=z

x + y + z = 2 z 0 m o d 2 \Rightarrow x+y+z=2z\equiv 0 \ \mod2

So the sum of numbers with same color is always divisible by 2 2 , and the sum of the numbers from 1 1 to 33 33 has to be even, but this leads to controversy, because: 1 + 2 + 3 + + 33 = 33 34 2 = 33 17 1 m o d 2 1+2+3+\dots+33=\dfrac{33*34}{2}=33*17\equiv1 \ \mod2

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