 So our path to the standard algorithm comes through an area that we might call division by partial quotients. So this is essentially just an improvement on our division by repeated subtraction, and we can improve the efficiency by removing large multiples of our divisor. So the problem is trying to figure out what these large multiples are can be somewhat difficult, but we can take the view point that any multiple we can find is something we can work with. And in fact, it's worth pointing out that we don't actually need to find the multiple that we need, we can just find some multiple that works, and a common strategy for doing this is known as chunking. The use of chunking leads to a method known as partial quotients. So let's think about a division like 3,010 divided by 86. So what we might do is we might subtract 86 over and over and over again, but that's going to take a long time. And it's worth noting that an easy multiple of 86 is times 10. 86 times 10 is 860, 86 times 100 is 8600. Now that's too big. We can't subtract this from 3,010, but we can certainly subtract 860. So what that says, too big so I can't subtract 100, but I can subtract at least 10. And if we're thinking about the order of magnitude of our answer, that says it's definitely less than 100. We have less than 186s in here. We have more than 10. So our answer should be somewhere between 10 and 100. Well, let's go ahead and start subtracting. So I'll start by subtracting groups of 1086s. And so what I'm going to do is I'm going to keep track of two things. One is I'm going to keep track of what I've subtracted and what I have left over. And the other thing is I'm going to keep track of how many 86s I've subtracted. So here this is 1086s, that gives me my 860. So by doing this subtraction, I've taken out 10 of these divisors. And let's see, I subtracted 860, oh, I can do that again. And I still have enough to do one more group of 86s. So now I've subtracted 861, 2, 3 times. I can't do that anymore. So I'm going to switch to just subtracting 86s. So I'll subtract once, twice, and a whole bunch more times. And I can do the subtraction this way. And here's the important thing. The bookkeeping part of this is how many 86s have I subtracted. So here's 10, 20, 30, 1, 2, 3, 4, 5, 35, 86s all together. And since I've subtracted my 3586s all together, my 3010 divided by 86, that's going to be 35 as my quotients. Well, did I have to subtract 10 at a time? Not at all. If I had recognized that I could have subtracted 30, 86s all at once, I could have done that as my first step. Now, recognizing that's a little bit more difficult, so let's see what we can do to get around that problem. So here's another example, divide 77, 85 by 1,773. So I might consider constructing a multiplication table for 173, and maybe I'll take advantage of an easy thing to do. I can just double the numbers. So here's 10, 173s give me that. If I double that, that's 20, 173s, that gives me that. If I double that again, that gives me 40, 173s, I have that amount. And if I double that again, I have 80, 173s, is that much? Oh, that's actually too much. I don't need this many 173s. I can get rid of it, but I don't really need to, I can just ignore it. But here's the important thing, I can definitely subtract 40, 173s immediately, because 40, 173s is a good deal less than what I'm starting with. So I'll subtract out about 40, 173s right away. I have 865 left. And the nice thing about this is I've already constructed a partial multiplication table for 173. Of course, these are the 10s, but the difference between the 10s and the 1s is just that extra 0 that I've added there. So I'll ignore the 0 for a second. And just using the work I've already done, I can see that I can subtract another 4 173s. So I'll subtract 4 173s, and there's 173 left over, so I'll subtract one more 173. And how much have I subtracted altogether? 40, 4, and 1. 45, 73s all together. And so that gives me my quotient, 7785 divided by 173. Quotient is 45.