sum of 3 out of 5 is divisible by 3

Problem:
Among 5 integers, there are always 3 with sum divisible by 3.
(From Problem Solving Strategies, Arthur Engel)

Solution:
Proof by Pigeon Hole Principle.
An integer is either 0, 1 or 2 \( \pmod 3 \). Imagine placing 5 integers into those 3 boxes. If we have at least one in each, then we can pick one from each, with sum divisible by 3.
Otherwise, we'll have at least 3 integers in one of the boxes. Pick those 3. QED

Comments

p - p^2-2 / p+2

J595 Mathematical Reflection

J595 Mathematical Reflection \(\sqrt[3]{(x-1)^2} - \sqrt[3]{2(x-5)^2} + \sqrt[3]{(x-7)^2} = \sqrt[3]{4x} \) By Titu Adreescu. My Solution (potentially wrong): Let \(A = \sqrt[3]{(x-1)^2}\) , \(B = \sqrt[3]{2(x-5)^2}\), \(C = \sqrt[3]{(x-7)^2} \), \(D = \sqrt[3]{4x}\). Hence we have \(A + C = B + D\). Notice that \(A^3 + C^3 = B^3 + D^3 \). Also notice that \( (A+C)^3 = (B+D)^3\) yields \( AC = BD \). Also by factoring out \(A^3 + C^3 = (A+C)(A^2-AC-C^2)\) and doing the same for \(B^3+D^3\) gives us \(A^2 + C^2 = B^2 + D^2\). Lemma 1 : \((A^3 - C^3)^2 = (B^3 - D^3)^2\). Proof: Note that \((A^3+C^3)^2 - (A^3-C^3)^2 = 4A^3C^3 = 4B^3D^3 = (B^3+D^3)^2 - (B^3-D^3)^2\) and the conclusion follows. \( \square \) Using Lemma 1 we get \( ( (x-1)^2 - (x-7)^2)^2 = (2(x-5)^2 - 4x)^2\) and solving for x gives us \( x = (9 \pm 4\sqrt{2}) , (3 \pm 2\sqrt{2}) \). \(\square\)

232 | 20^n + 16^n - 3^n - 1

Maths by Chance

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