 Alright, so let's look at the long division algorithm. So let's take a typical problem, 6,915 divided by 17. So you know how this works. We're going to set down our dividend divisor, 6,915 divided by 17. And we begin with this process, 69 divided by 17 is going to be what? Well, it's around 3, and I'll write it here. Wait a minute, can I write it there? No, I have to write it someplace else. Well, let's see what happens. What am I going to do? Well, I'm going to go 3 times 17, that's going to be 51. Well, now really 3 times 17 is 51, but I shouldn't write the 51 here. I should be writing it way over here. And so what we're really doing is we're not multiplying 3 times 17 to get 51. We're actually multiplying 300 times 17 to get 5,100. And that's what allows me to subtract here to get 1,815 left over. Well, let's do that again. 17 into 18, well that goes about once. And again, 1 times 17 is 17, but again, I'm not actually getting a 17 here. I'm actually getting a 1,700. And to get that, I need to go 100 times 17 to get me my 1,700. And now I can do my subtraction to have 115 left over. 17 into 11 doesn't work. 17 into 115, well guess about 5 times or so. And it doesn't really matter where I write it. 5 times 17 works out to be 85. And now I can just do the straight subtraction to get me 30. 17 into 30 only goes one time. One time 17 is 17, subtract, and I get my remainder 13. So now remember, this division is how many 17s can you subtract? How many 17s do you have to subtract until you have a remainder that you can't subtract a 17 from? So what did we do? Well, we started out by subtracting 317s. Then we subtracted 117s. And then we subtracted 5. And then we subtracted 1. So altogether I've subtracted 300, 400, 405, 406. So I've subtracted 406, 17s altogether. And what's left at the end of it? Just 13. So my quotient is 406 with remainder 13. And all that we've done here, really, is just partial quotients. And the only change that we've made is we've written down our quotients in a different place. Instead of keeping track of them on the side, we're keeping track of them up above. Alright, well, now let's suppose that we had a magical gift and we could guess perfectly without making any mistakes whatsoever, ever. And so if I could guess the partial quotients perfectly without hesitation every single time, I might make this go a little bit more efficiently the following way. 17 into the first two digits. Well, that goes 400 times. And 400 times 17 is 6,800. Get rid of that. I have 115 left over. 17 into 150, if I make a perfect guess, that's going to go 6 times. 6 times 17 is 102. And I have 13 left over. And my perfect guessing scenario, if I could guess perfectly without hesitation every single time I do this problem, then I can use the method of partial quotients very efficiently. And I can read off my answer. 406, remainder 13. Alright, well, let's suppose we want to play around with this a little bit. Suppose I wanted to line up these partial quotients with the dividend. Why would I want to do that? I don't know, but suppose I want to, just because I think it looks pretty. So this 400 should actually go a little bit further over. And the 6 actually needs to go in the ones place. Now, it's worth noting I have hundreds, I have ones, but I have no tens. And so just because I feel like it, maybe I'll indicate that I have zero tens by putting in the tens there. And to get the standard algorithm, I'm going to drop all of these extra zeros that I don't really need to use. And so there's my tremendous saving in space. Well, maybe not so much, but I've saved a lot of really expensive ink. And I'm not writing those zeros, and there's my quotient using the standard algorithm. Hardly seems worth it. Alright, so let's take a look at another division problem. How about 8,915 divided by 26. This time we'll make a comparison of the two methods. So I'll set it by quotient 26 into 89, I guess perfectly without hesitation. Every time that's 3, and in order to get that 3 times 26 is 78, but in order to get it all the way over here, I need to multiply, in fact, by 300. And I'll subtract, and here's what I have left. 26 into 111, I guess perfectly, and without hesitation, 4. 4 times 26, 104. And again, to get it into this place, I actually need that to be a 410. So that's going to be 40. I'll multiply and subtract. And again, I guess the quotient without hesitation and perfectly, 26 into 75 goes 2 times, and 2 times 26, 52, and subtract, and I'm left with 23. 26 is larger than 23, so I'm now done with the problem, and my quotient 342 with remainder 23. And so there's my division problem done using partial quotients. Now, if I want to use the standard algorithm, what I'm going to do is I'm going to shift all of my partial quotients so they line up with the dividend. Again, not obvious why you would care, but we'll go ahead and shift them so that they line up with the dividend. There are these zeros that we didn't really need, that we had to write down when we did this method using partial quotients, and then one last thing, we'll collapse this partial quotient so that it's written on one line. And so we have a nice comparison of the two methods. Over here, we have our method of partial quotients, and here we have our standard algorithm. And as you can see, the standard algorithm is more compact. We've avoided having to write down a couple of zeros, and we arrive at the same answer that we got to with the partial quotients method.