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.
To solve this problem, we need to find the least positive integer "n" that satisfies a specific condition related to coloring a set of the first "n" positive integers. Let's break down the problem step by step.
Step 1: Understanding the Problem:
We have a set of the first "n" positive integers, {1,2,3,...,n}, and we can color each number either red or blue. The goal is to find the smallest possible "n" such that no matter how we color the set, there will always exist four integers with the same color (red or blue) that satisfy the equation x + y + z = w.
Step 2: Analyzing the Problem:
To tackle this problem, we will employ a technique called the Pigeonhole Principle. This principle states that if more objects are put into fewer containers, then there must be at least one container with more than one object.
In our context, the containers are the possible sums (x + y + z) that we can form from three elements in our set, and the objects are the different ways we can color the set. The objective is to find the smallest "n" satisfying the condition, meaning we aim to minimize the number of containers.
Step 3: Finding the Solution:
Let's simplify the problem by considering the largest possible value for x, y, and z to minimize the number of containers. We will set x = n, y = n-1, and z = n-2. The sum of these values is:
(n) + (n-1) + (n-2) = 3n - 3.
Therefore, the minimum value for "n" should have at least 3n - 2 elements in the set. Now, let's consider the coloring possibilities.
If the set has 3n - 2 elements, then there are 2^(3n - 2) ways to color it because we have two choices (red or blue) for each of the 3n - 2 elements.
However, we need to consider one additional case. If all the sums (x + y + z) formed by any combination of three elements are distinct, we could get up to C(3n - 2, 3) distinct sums. Since we have 2^(3n - 2) possible colorings, we want to find the smallest "n" such that C(3n - 2, 3) > 2^(3n - 2).
Step 4: Checking each "n":
Let's check the values of "n" starting from 1 and see which one satisfies the condition:
For n = 1:
C(3(1) - 2, 3) = C(1, 3) = 0, and 2^(3(1) - 2) = 2^1 = 2.
The condition C(3(1) - 2, 3) > 2^(3(1) - 2) is not met.
For n = 2:
C(3(2) - 2, 3) = C(4, 3) = 4, and 2^(3(2) - 2) = 2^4 = 16.
The condition C(3(2) - 2, 3) > 2^(3(2) - 2) is not met.
For n = 3:
C(3(3) - 2, 3) = C(7, 3) = 35, and 2^(3(3) - 2) = 2^7 = 128.
The condition C(3(3) - 2, 3) > 2^(3(3) - 2) is met.
Therefore, the least positive integer "n" that satisfies the given condition is 3.
Step 5: Answer:
The smallest possible value for "n" is 3. This means, regardless of how the elements of the set {1,2,3} are colored (red or blue), we can always find four integers with the same color that satisfy the equation x + y + z = w.