 In this video, we'll introduce a few common strategies for using the Pigeon-hole principle. To show a strategy for developing proofs, we'll rely on two main themes. Show that something will eventually occur, and then find a bound on when it must occur. To see how this might work, let's consider the problem. Let k be a positive integer. Prove that in any sequence of k integers, there is a set of consecutive integers whose sum is divisible by k. And this is something you might see in a problem set. Now, while we could solve the problem, let's change it so we incorporate our two main themes. The problem gives us the bound, so let's omit that, and then try to find it on our own. The problem also tells us that it will occur, since we wouldn't be trying to prove it otherwise, so let's leave that open. This makes the problem harder, so let's make it a little easier by considering a specific case. So a rewritten problem might look something like this. So here we just have a sequence of integers, and we don't know if there will be a sequence of consecutive terms whose sum is divisible by 17, but if there is, when will it occur by? So in general, organize. There could be many possible sums of consecutive integers like, or, or, so how can we organize them? One possibility is just to consider the partial sums. S1 is the sum of the first term, S2 is the sum of the first two terms, S3 is the sum of the first three terms, and so on. But it's best to avoid the infinite, so let's cap the partial sums at some point, and remember, you can assume anything you want as long as you make an explicit. So let's take n to be a large number, and we'll go back and fix our assumptions later on. So we'll leave the details of n alone until later on. Now since we're looking for a sum that's divisible by 17, let's consider the remainders when these partial sums are divided by 17. The remainders when the Si are divided by 17 are among the values. Now to apply the pigeonhole principle, we want the number of items to exceed the bins. So if we make the remainders the bins and take n greater than 17, we're guaranteed at least two will have the same remainders. So we'll fix our assumption. Now a good habit to get into is never erase, cross out what you don't want. So here n isn't just a large number, it's actually n is greater than 17. And since our n greater than 17 sums, and only 17 possible remainders, at least two of the sums have the same remainder. So remember, definitions are the whole of mathematics, all else is commentary. We're saying that two sums have the same remainder. What does that mean? Well it means our two sums are 17 times a quotient plus a remainder, and 17 times a different quotient plus the same remainder. So we can subtract. And again, definitions are the whole of mathematics, all else is commentary. And this includes any definitions we happen to introduce ourselves. Remember the s's were our partial sums. So we may assume that i is greater than j. So that means Si will have all terms up to j, and then some more. And if we subtract sj we get. And so this gives us our sum of consecutive terms, divisible by 17. And finally to complete the process, we note that we've solved a specific problem. In any sequence of more than 17 integers, there is a set of consecutive terms, whose sum is divisible by 17. And the proof generalizes for any divisor. So this actually gives us a theorem. Let Q be any positive integer. Any sequence of more than Q integers contains a set of consecutive terms, whose sum is divisible by Q. And that is the problem we originally started with.