Percentage of Numbers Divisible by 6, 8
Date: 03/18/2003 at 20:45:24 From: Mark Subject: Divisibility by 6, by 8 What percentage of numbers is divisible by 6? What percentage of numbers is divisible by 8? I really don't understand the principle of the question. For instance, what numbers?
Date: 03/19/2003 at 08:51:18 From: Doctor Rick Subject: Re: Divisibility by 6, by 8 Hi, Mark. The answer does depend on "what numbers," but maybe not as much as you'd expect. If it means all real numbers, the answer is zero percent, because only integers can possibly be divisible by 6, and the ratio of integers to real numbers is zero. We say that the integers constitute a set of measure zero. You can read a bit about this here: Countable Sets and Measure Zero http://mathforum.org/library/drmath/view/51860.html The connection between this notion and probability (which is related to your "percentage" question) can be seen here: Question on Probability of Repeating Digits http://mathforum.org/library/drmath/view/52212.html But let's get on to what is really in view in this problem. What if "numbers" means "integers"? A non-zero percentage of all integers is divisible by 6, even though the "part" (the number on top in your percentage calculation) and the "whole" (the number on the bottom) are both infinite. How can this be? Consider any finite contiguous set of integers, let's just say from 1 through N. Then there are N numbers in the "whole." How many are there in the "part"? Does the percentage depend on N? Just slightly, and that dependence gets smaller as N increases. If you take the limit as N goes to infinity (so that you're considering all whole numbers), you'll have a specific value for the percentage. That's an outline of what's involved. You'll have to decide how much detail you need to go into, depending on what sort of class this is for. To me the answer is intuitively obvious, but to prove it rigorously would take more work. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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