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Non-Dividing SetsSubmitted by Geoff, 30 July 2002. Original answer and this article by Allen Stenger.I am asked to find the largest number of elements that a set of integers from 1 through 100 can have so that no one element in the set is divisible by another. I was told a "hint:" Imagine all the numbers 1--100 in the form (2^k) * m where k ≥ 0 and m is odd. I have DONE this for nearly all the integers 1--100 but cannot see a viable pattern. I don`t want the answer to this question just help interpreting the hint or maybe some insight into a nother way to solve it. Any help would be great. Thanks! Hint 1Work on the following more general problem: Among the set of integers from 1 through 2n, what is the largest subset such that no element divides another element? The reason this is a better problem is that
Hint 2Experimentally the top half of the numbers is such a set; that is, no one of these numbers divides another: A = {n + 1, n + 2, ..., 2n} But adding any number to this set loses the property, because any additional number k is in the range 1, ..., n and therefore divides the 2k that is already in the set. Can you prove the following two statements?
Hint 3Part (1) is easy. Let a and b be two elements of A , with a < b . Can b/a be an integer? No, because it is greater than 1 and less than 2! Here's another hint for the toughie. The example of adding an element to A suggests that in this problem there is some special relation between a number and its double. If we have a set B with more than n elements, group the elements into subsets such that a number and its double go together if both are in B . More specifically, for each odd number m define the subset Bm as all numbers of B of the form 2^k * m where k ≥ 0. The Rest of the SolutionIf one of the Bm contains more than one element, we are done, because the smaller element divides the larger one (in the quotient of the two elements, the m cancels out, and the smaller power of 2 divides the larger power of 2). So we want to show there is a Bm with more than one element. And this is easy, because there are n such Bm (one for each odd number 1, 3, ..., 2n - 1 ), while there are at least n + 1 elements in B . The pigeonhole principle tells us that some Bm has more than one element. The Pigeonhole PrincipleWe used a simple but very important idea called the pigeonhole principle (also called the box principle, drawer principle, or in German Schubfachprinzip). It is attributed to Peter Gustav Lejeune Dirichlet (1805--1859). (Dirichlet was a German with a French-sounding name, pronounced dee-ree-CLAY.) Dirichlet used the principle in number theory problems. It has several equivalent statements, such as:
There are even infinite versions:
The pigeonhole principle is useful in many number theory and combinatorics problems. Here are a couple of problems for you to practice on:
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