Determine the least positive integer n for which the following condition holds: No matter how the elements of the set of the first n positive integers, i.e. {1,2,…n}, are colored in red or blue, there are (not necessarily distinct) integers x,y,z, and w in a set of the same color such that x+y+z=w.
Details and assumptions:
The phrase not necessarily distinct means that the integers can be repeated. For example, if 1,2,4 are all colored red, then we have 1+1+2=4 which would satisfy the condition.
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To find the least positive integer n for which the given condition holds, we can use a systematic approach. We start by considering small values of n and gradually increase it until we find the smallest value that satisfies the condition.
Let's start with n = 1. In this case, the set of positive integers is {1}. However, we cannot form any valid equation with x, y, z, and w using only one element. Therefore, n = 1 does not satisfy the condition.
Moving on to n = 2, the set of positive integers is {1, 2}. We can color them as either red or blue. The possible equations we can form are:
- Red: 1 + 1 + 1 = 2 (1+1+1=3)
- Red: 1 + 2 + 1 = 3
- Red: 2 + 1 + 1 = 3
- Blue: 1 + 1 + 2 = 4
As we can see, there is at least one valid equation for each coloring option. Therefore, n = 2 satisfies the condition.
Now, let's check n = 3. The set of positive integers is {1, 2, 3}. We can color them as red or blue.
The possible equations we can form are:
- Red: 1 + 1 + 1 = 3
- Red: 1 + 2 + 1 = 4
- Red: 1 + 1 + 2 = 4
- Red: 2 + 1 + 1 = 4
- Red: 2 + 2 + 1 = 5
- Red: 2 + 1 + 2 = 5
- Red: 3 + 1 + 1 = 5
- Blue: 1 + 1 + 3 = 5
- Blue: 1 + 2 + 2 = 5
- Blue: 2 + 1 + 2 = 5
Again, we see that there is at least one valid equation for each coloring option. Therefore, n = 3 satisfies the condition.
Continuing this process, we find that for n = 4, one coloring option (red) allows for a valid equation, but the other coloring option (blue) does not. Therefore, n = 4 does not satisfy the condition.
Hence, the least positive integer n for which the given condition holds is n = 3.