 Alright, so let's take a look at 50 ways that we can do a subtraction. And again, we're not actually going to show 50 ways, but we'll show a couple of basic ways with lots and lots of possible variations. So consider this problem, subtract 317 minus 85, and let's try to show at least five distinct methods. So let's start by counting up, and this is going to be a count up from 85 to 317, and we'll see how far we have to go. So I'll start at 85, and I'll count up plus 5 to 90 to get to a convenient number, plus 10 to 100 to get to our next convenient number, and then we'll take a couple of larger steps. So now I want to get to 300. I'm getting close to my target, so I have to slow down a little bit. Maybe I'll add 17 at that point, and now I can see how far I've counted up. That's 200, 210, 227, 232. So my subtraction is going to give me 317 minus 85 is equal to 232. The other thing I might do is I might try counting down from 317 down to 85. So let's go ahead and take that. So we'll start at 317. I want to get down to 85, so I'll take a large chunk at the beginning. I'm going to subtract 200, and let's see. I'll slow down a little bit. I'll subtract 17 to get me to 100, and then I'll drop 10 to get me down to 90, and then 5 more to get me down to 85. And all together, how far have I counted down? 200, 210, 227, 5 more, 232. So again, 317 minus 85 gives me my remainder, 232. Now, again, we don't have to count down in exactly this fashion. We can consider other variations. So for example, 317, we might go down 17 to 300, and then down 15 to 285, because I know I want to get to 85, and then I'll subtract 200 at that point to take me all the way down. So there's 200 plus 15 plus 17. So here I might add this as 32, and 200, again, 232 is my difference. All right, so method three, let's try counting past. So here I'm removing 85. Well, if I remove too much, I have counted past, and then I've got to return some amount. So in this case, well, I might notice that 100 is a little bit more than 85. So I'll subtract 100, which is too much, and then I'll return the 15 I shouldn't have subtracted. So starting at 317, I'll remove 100. That takes me down to 217, and, well, that's too much. So I've got to return 15, which takes me back to 232. All right, how about method four? Let's try equal add-ins. So in this case, what we're doing is we're doing the subtraction 317 minus 85. And in this case, I note that, again, 85 is just a little short of an easy number to work with, 100. So I'll add 15 to it to get 100, and I have to do the same thing to the original number, 317, add 15, takes me up to 332. And what makes this convenient is I can do this subtraction pretty easily. 332 minus 100 is just 232, and that 332 is the same as the answer to the original question. All right, so that's four methods. Let's go to a fifth method. Let's write decomposition. So here I'm going to decompose the subtrahend, and 85, if you speak it out, is 80 and 5. So I can subtract by decomposing 85 into 80 and 5. So I'll subtract 80 to take me down to 237, subtract 5 to take me down to 232 as my subtraction. Okay, so that gave us five different methods. We had counting up, counting down, counting past, equal add ends, and decomposition. Now, as a sixth method, we could also do this using these standard algorithms. So we'll subtract 317 minus 85, and we'll use our standard algorithm. So there's our subtraction, and we can go through the process. 5 from 7 gives us 2, 8 from 1 can't do that, so we need to unbundle and trade. So we're going to take one of these units here, drop that, and then add 10 to the next. That's 11, and 8 can be subtracted from 11. That gets us 3, and then 2 brings straight down to 232.