 Alright, so let's put our multiplication methods together and find 50 ways to multiply two numbers. Now again, we won't necessarily get up to 50. We don't have that many distinct methods, but with variations on how we apply a method, we can easily find not just 50, but maybe 500 ways of multiplying two numbers. But we'll pick out a few that we'll want to work with. So let's take 135 times 48, and we want to show this multiplication using five different methods. So the first observation I might make here is that 48 is actually a number that has a whole bunch of different factors. So maybe I want to do multiplication by factors and using the associative property. So let's see, 48 is 4 times 12, so that's a good place to start. And you know, I really don't feel like doing a lot of work. It's a nice sunny day, and so I really don't want to put a lot of effort into doing this multiplication. So let's go ahead and break these up as much as possible. So 4, that's 2 times 2, and 12 is 3 times 4, and 4 is 2 times 2. And so 48 is this product of a whole bunch of terms, and the key here is that multiplying by 2, easy, multiplying by 2, easy, multiplying by 3, a little bit harder, but really not that difficult, and so on. So here I've broken this 48, which is a two-digit number, which would be complicated to multiply by, and I'm going to multiply it out using a whole bunch of little steps, because I don't feel like thinking a lot today. And so I'll show this multiplication using the arrow diagram. So there's my 135 times 2 gets me 270 times 2. Let's see, that's 200 times 2 is 400, 140, 540 times 3, a little bit more work. That's 1,500 plus 120, that's 1,620 times 2 again, 3,240 times 2 last times 6,480. And so there's my multiplication, lots of little, easy steps, and I end up with my product, and this is because I don't really feel like doing a hard multiplication. The price we pay is we do a lot of little steps. And again, worth noting, I don't have to apply the multiplication by factors in this exact fashion. I can choose any set of factors that I want to, as long as I get 48, and I can also do the factors, do the multiplications in any order that I want, so maybe I decide times 3 is the hard product, so I'll do that first to get that out of the way. Times 3, that's 135, thinking the distributive property, that's 390 and 15. That's 405, and then by times 2, you'll note, become a lot easier to do. 405 times 2 is 810, times 2, 1620, times 2, 3240, times 2, 6480. And again, to really see why this is advantageous, do the problem in your head. Do the problem without writing anything down. This is easy to do mentally. It's a lot harder to do this, without having to write anything down. And again, one point worth noting, this is still, both of these are the same method, their multiplication by factors, so these don't count as distinct methods. They are different ways of applying the same method. But a different method, we could use decomposition, which is to fall back on the distributive property. Again, if you're doing this problem mentally, you might use the decomposition strategy, and possibly in combination with the factor strategy. So we can decompose 48 into 40 and 8, and applying the distributive property, that 48 splits 40 and 8. And I can do the multiplication, 135 times 40, 135 times 8. And again, my next step, well, I might say that 48 is 4 times 10, and that's 2 times 2 times 10. So I can do this product, 135 times 40, by doubling twice, and then times 10. That's 270, 540 times 10 is 5400. And here, I double three times, 135, 270, 540, 1,080. So that's 5400 plus 1,080. And now I can just add these two numbers, that's going to be 6480 as my product. Now, here's another way I might approach it. I could use the distributive property in another way, using the fact that 48 is 50 minus 2. Now, as a note here, as far as a different method, we're actually relying on a different property of our numbers. So in the previous method, we used the fact that 48 was 40 and 8. Here we're using the fact that, which is an addition, here we're using a different property, which is that I can express 48 by a subtraction. So this would actually count as a different method. So I'll do this multiplication, 135 times 48, 135 times 50 minus 2. And I can expand that out, 135 times 50, 135 times 2. And here I might want to do this multiplication. That's 50 times, break this up, 130 and 5. So 50 times 100 is 5,050 times 30 is 1,500, and 50 times 5 is 250. And in this one, that's pretty easy to do. That's just 135 times 2 is 270. So this first product expands 5,500, 250 minus 270. And I'll add these together, 6,750 minus 270. And if I want to do the subtraction, maybe I'll subtract past. So that's minus 300 return 30, so that gets me to 6,480. And again, note that we're using all of our previous methods, all of our knowledge here to make this multiplication a lot easier. Because again, so again subtracting 270, well, we could go off here and do the standard algorithm, but really it's minus 300 return 30 gets us to this point. All right, so that's three methods, how about an area model. So we're going to multiply 135 times 48. Let's go ahead and draw an area model for the product. That's going to be a rectangle, 48 by 135. And again, the key to the area model is I'll split the factors into pieces I want to work with. So this 135, maybe I'll split that into 135 and this 48, I'll split that into 40 and 8. And now I have a whole bunch of rectangles and I can find the areas pretty easily. That's 100 by 40 is 4,000 and 100 by 8, 800, and so on. So my product is going to be the sum 4,800, 1,440, 240, and I can add those two together to get 6,480 as my product. Well, how about as a repeated addition? So again, this is the product 135 times 48, well that's 135,48. And so I might do that as the following. So I have 148, well that's 4,800. I have 30, 48, that's 1,440. And I have 5, 48, and that's going to be 240. And I'll add those together to get my total 6,480. So that gives me five different ways of doing the multiplication. I can do repeated addition. I could use an area model. I could use the distributive property this time using a subtraction. I could use the distributive property and decompose one of the factors. Or I can use the associative property and multiply by factors. As a sixth method, we could consider what the standard algorithm looks like. So this 135 times 48, I'll multiply this using the standard algorithm, although I'll write it out in some detail first. So if I use the standard algorithm, that's 8 times 5 is actually 40. 8 times 30, 240. 8 times 100 is 800. 40 times 5, 200. 40 times 30, 1,200. 40 times 100 is 4,000. And it's worth comparing these products here, 240, 800, 200, and so on. It's worth comparing those products to the numbers you get in the area model. Notice that these partial products here are exactly the same numbers that we got in the standard algorithm. And so what's the last thing we do? Well, the sum is the product. So we'll add up these numbers and get our product. Okay, well, this is not quite the standard algorithm. We get the standard algorithm by doing a couple of things. First of all, it's really tedious and difficult to write these extra zeros because that takes a whole third of a second to write them out. So we'll drop those zeros out and we'll smash all of these partial products together to form our sums. And here's our standard algorithm. Here's our somewhat expanded standard algorithm. And then there's five other ways we could have done this, multiplicating. What's the best way of doing it? Take your pick. It doesn't make a difference. What is most efficient is going to depend on what the problem is, what you have to work with, and how the question is being asked. If I had asked the problem, find this product mentally without writing anything down, you're probably not going to picture this in your head. You're probably going to use some variation of multiplication by factors or by the distributive property. If you have the space to write something down, maybe you'll do this, maybe you'll do this, maybe you'll write an area model. If the numbers were different, 99 times 97, then other methods will be more efficient. And so the question is not what's the best way of solving a problem, but what's the best way of solving this problem as it's written, as it's presented to you, how you feel this day, and with the materials you have available.