 So by now, we've looked at a number of different ways of adding, so let's see how many we can come up with for a single problem. We might not actually get up to 50 ways because we're including variations on that, but let's see if we can do at least five different methods. So how shall we start? Well, probably the easiest we can start with counting on, and so that means we're going to count up from 57 up to 538. So we'll start at 538. We want to count up by 57, so well, let's start by counting up 50, and that takes us to 588. Let's be a little bit creative here, or lazy as the case may be, and if I count up from 588 to more, I get to a number that's easy to work with, and that's taken me up 52, so I need five more to get my total of 57. So counting up 538 up to 595, there's my total. Now, we say that there's 50 different ways of doing this, but here is something that's actually not a different method, but it's the same method done slightly differently. So rather than counting up from 538, 50, and 2, and 5, we might count up 538 up to an easy number to work with 540, and then maybe up another 50 to 590, and again, here I have 2 and 50, that's 52. I need five more to get to my final answer. And again, this is still counting on, so this does not really count as a different method. It's the same as the preceding method is just being done in a different way. However, a distinctly different method is through decomposition. So let's decompose our add ends and add that way. So what are we going to do? We'll take our sum, 538 plus 57, and I'll break my add ends up. So 538 is going to split up into 538. Likewise, 57 is going to be broken apart into 57. And now that I've decomposed the two add ends, I can add by the pieces. This is 500, 30 and 50 is 80, 8 and 7 is 15, and finally, we'll glue the pieces back together. 580, 15 is 10 and 5. That's 580, 590, 5 as my final answer. Another way we could do this problem is through partial add ends. I don't have to add 57 all at once. I can add it in pieces. So we'll take our first add end, 538, and I'll add 50, and that gets me up to 588, and then I'll add 7, that gets me up to 595, and altogether I've added 57, which is what I wanted to do. Well, what else can we do? Well, we might notice that 57 is almost 60. So counting past also works as a possible strategy for this. So in this case, I'm going to count past where I need to go. 538 plus 57, well, if I add 60, that's a little bit too much, and I'll have to work my way backwards. So I can show that as that follows. So I'll take my 538, I'll add 60. Oh, that's too far. Got to go back three units back to 595. Well, let's look at something that looks a little bit more like the standard algorithm, except we'll do this as totals below. And so again, I'll add the numbers 538 plus 57, and the totals below method, I'm just going to add within the column. So 5 and whatever this is, nothing there is going to give me 5. 3 and 5 gives me 8. 8 and 7 gives me 15, which I'll write this way. There's my totals below. And now I'll add a second time. There's 5 and 9 and 5 as my final answer. Now you might notice that none of these methods that we've used is the standard algorithm. We have totals below. We have counting past. We had partial add ends. We had decomposition. And we had counting on five distinct ways that look nothing like the standard algorithm. So let's take a look at the standard algorithm. That'll actually be our sixth method. So we'll go ahead and set up the addition using the standard algorithm. So there's our numbers written in the column format, one above the other. And so what do we have to do with our standard algorithm? We only have to do 8 plus 7 is 15, but I can only write down the 5. I have to write the 1 over here someplace. And then I have to add these numbers. 1 and 3 and 5 is 9. I'll set that down. And then the 5 here gets carried down to 595. Now it's worth pointing out that the main advantage to the standard algorithm is it's compact. It saves space on the printed page. However, the question is, have we paid too much for the compactness? Paper cost us about a tenth of a cent per page. So it might not be worth saving the money if we lose something in efficiency. And so the natural question to ask is, of these methods, what's the most efficient? And this goes back to the question of adaptive expertise. If you have adaptive expertise, you should be able to look at a problem like this and identify the best ways to solve the problem as it's presented to you on this device today. With the materials you have readily available, under the conditions that you're going to solve it. But if the problem changes, your best and most efficient method will likely change as well. For example, let's change the problem a little bit. Add 538 and 57. Now you might say, well, that's the same problem, but it isn't. In this case, you're not given the digit form of the add-ins. You're given a verbal expression for the add-ins. And in this case, some approach like counting on might work more efficiently. Or let's add a restriction. Suppose we don't allow you to use pencil and paper. You can't write anything down, but you have to do this problem mentally. And so we have the same problem, 538 plus 57, but this time we have to do the problem in our heads. And again, the method that we use to solve this problem could be very different from the method that we use to solve this problem that we can use pencil and paper.