Level C Lesson 9 Multiplication of Whole Numbers with Property Application

Level C Lesson 9
Multiplication of Whole Numbers with Property Application

In lesson 9 the objective is, the student will apply properties of operations as strategies to multiply. The student will multiply one-digit whole numbers by multiples of 10 using strategies based on place value and properties of operations.

The skills students should have in order to help them in this lesson include multiplication facts for tables 0 through 9.

We will have three essential questions that will be guiding our lesson. Number 1, how can I determine what happens when any single-digit number is multiplied by 10? Number 2, how can I use different strategies to multiply two factors? And number 3, how can I use different strategies to multiply three factors?

The SOLVE problem for this lesson is, Marcel’s mother is buying gum for his soccer team. The cost of gum is $2.39 per package. She buys 3 packages of gum. Each package has 5 packs of gum. Each pack of gum has 10 pieces in it. How many pieces of gum are in all 3 packages?

We want to start by Studying the Problem. First we want to identify the question within the problem. When we identify the question, we want to underline the question. How many pieces of gum are in all 3 packages? Next we will put this question into our own words in the form of a statement. This problem is asking me to find the number of pieces of gum in all three packages.

During this lesson we will learn how to multiply whole numbers with property application. We will use this knowledge to complete this SOLVE problem at the end of the lesson.

Throughout this lesson students will be working together in cooperative pairs. All students should know their role as either partner A or partner B before beginning this lesson.

We will start this lesson by building multiplication arrays using the grid that is provided and our beans. On grid A we want to show the fact 3 times 10 or 3 groups of 10 items. Since each group has 10 items we will need to place 10 beans on each row. We will have a total of 3 rows of 10 beans to represent our 3 groups of 10 items. We have now created an array representing 3 times 10. 3 and 10 are the factors that we used to create this array. The factors that we will use to create an array on grid B are 10 and 3. We want to create an array that shows 10 groups of 3 items. Each row needs to contain 3 beans to represent a group of 3. We will have a total of 10 rows to represent our 10 groups. Grid a represents 3 groups of 10 items, and grid B represents 10 groups of 3 items. The arrays have the same number of items but are built differently as the first array has 3 groups with 10 items in each group. And the second array has 10 groups with 3 items in each group. Both arrays show a total of 30 beans.

Now let’s take our concrete representation in using the beans and change it into a pictorial representation, replacing each bean by shading in the grided square where the bean was. We will do this for both grid A and grid B. Now that we have finished shading in each array on grid A and grid B let’s answer a few questions about the grids. What does the first grid represent? 3 groups with 10 items in each group. What is the number fact for grid A? 3 times 10 equals 30; and we will label our array on grid A. Now let’s look at grid B. What does the second grid represent? 10 groups with 3 items in each group. What is the number fact? 10 times 3 equals 30; and we will label this fact on grid B. Remember that although our arrays look different they both have products of 30. Now let’s compare the grids. The arrays have the same number of items, but are built differently. The first array has 3 groups of 10 items, and the second array has 10 groups of 3 items. 3 times 10 equals 10 times 3.

Let’s take a look at another example using the grids that are provided. Let’s start, by making a prediction as to what the products represented, by the 2 arrays are going to be. Looking at the 2 arrays I’d say that the products, are going to be equal. Looking at grid A, what does this mean? We have 4 groups because we have 4 rows that are shaded in and each row has 10 grided squares that are shaded. So here we have 4 groups of 10 items. There are a total of 40 squares that are shaded. Are fact is 4 times 10 equals 40. Now, let’s look at grid B. What does the array in grid B mean? Our first group has 20 shaded squares, and our second group has 20 shaded squares. So the fact in grid B is 2 groups of 20 items or 2 times 20. There are a total of 40 shaded squares, so our fact for grid B is 2 times 20 equals 40. Our prediction that the value of grid A and grid B are the same is correct. Because 4 times 10 is equal to 2 times 20. The arrays have the same product but are shown differently. Both in the last example that we completed, and in this example, we are multiplying a single digit, by a multiple of 10. What happens when you multiply any single digit by a multiple of 10? You multiply the value of the single digit by the value in the tens place and add a zero. Notice if we take our single digit in 4 times 10, the single digit will be 4 and if we multiply it by the value in the tens place which is 1 we get 4 and then add a 0 in the ones place we get 40. Which was our answer on grid A, on grid B our single digit value is 2 and the value in the tens place of the number 20 is 2. So when we multiply 2 times 2 we got 4, and when we added a 0 in the ones place we got 40.

Let’s try this with another example, 30 times 5. Our single digit is 5 and the value in the tens place of our 2 digit number is 3. 3 times 5 equals 15. So when we multiply 30 times 5 we have 15 and we add a 0 in the ones place, so 30 times 5 equals 150. By using this rule we were able to complete this example without having our grid with the pictorial model on it to help us.

Now let’s look at a different type of problem. The array on grid A means that we have 5 groups with 6 items in each group. We can label this fact on our array as 5 times 6 which is 30. On grid B we are representing 5 groups with 4 items in each group in the first array, and 5 groups with 2 items in each group on the second array. Our first array is 5 times 4, and our second array is 5 times 2. In grid B we need to add our 2 arrays together. Even though the array in grid A looks different from the arrays that we have in grid B, we know that both sets of arrays are equal, because they both have the same number of squares that are shaded. There are a total of 30 shaded squares in grid A, and there are a total of 30 shaded squares in grid B. So the array in grid A is equal to the 2 arrays in grid B. Let’s take a minute to talk about why that is. On grid A we have 5 groups of 6 items. And on grid B we have 5 groups of 4 items plus 5 groups of 2 items. Grid A has 6 items in each group and grid B has 4 items in each group on the first array and 2 items in each group on the second array. Notice that our 6 items in each group on grid A are equal to our 4 items in each group plus our 2 items in each group on grid B. If we add our 4 items in each group and our 2 items in each group together on grid B it is equal to the 6 items in each group on grid A. What has happened is that our 6 items in grid A have been broken apart into 2 separate arrays on grid B. This breaking apart or distributing of a factor is also known as the Distributive Property.

Let’s look at another set of grids. Grid A in this example means 6 groups with 7 items in each group. We write this as 6 times 7 which equals 42 total items. There are 42 shaded squares on grid A. Grid B represents 6 groups with 5 items in each group and 6 groups with 2 items in each group. Our first array can be represented with 6 times 5, and our second array can be represented by 6 times 2. These arrays are together on grid B, so we add these 2 grids. Grid A equals 42 and grid B equals 42, as there are 42 shaded squares on each grid. So even though they look different grid A is equal to grid B. How can we tell that the array on grid A of 6 times 7 is equal to the 2 arrays on grid B which are 6 times 5 plus 6 times 2. We can do this by combining the arrays from grid B on grid C. We will draw a picture on this third grid to show the combined arrays. Right now we have the array of 6 groups with 5 items in each group, and the array of 6 groups with 2 items in each group. When the arrays are pushed together the new array becomes 6 times 7. The 2 arrays from grid B were added together. The 5 items from each group from the first array and the 2 items from each group from the second array when added together give us a total of 7 items in each group on grid A.

Let’s talk about each array on grid A and grid B represents in this next example. On grid A each array shows 2 groups with 3 items in each group or 2 times 3. There are a total of 4 arrays on grid A. On grid B each array has 3 groups with 4 items in each group. We represent this with 3 times 4. And there are 2 arrays on grid B. We will place parenthesis around the 2 and 3 on grid A, and we will place parenthesis around the 3 and 4 on grid B. By looking at the arrays on grid A and the arrays on grid B we can see that the 2 grids are equal. They both have 24 squares that are shaded or a product of 24. We can write the facts represented by the arrays on grid A as the quantity 2 times 3 times 4 total arrays. We can write the fact to represent the arrays on grid B as the quantity 3 times 4 multiplied by the 2 arrays on grid B. There are 24 shaded squares on grid A, and there are 24 shaded squares on grid B. So both grids have a product of 24.

The next set of grids, grid A shows 5 groups with 2 items in each group or 5 times 2. There are 3 arrays on grid A. We can represent the facts shown on grid A as the quantity 5 times 2, which is what each array represents, times a total of 3 arrays. On grid B we have 5 groups with 3 items in each group represented in each of our arrays. We write this as 5 times 3. There are a total of 2 arrays on grid B, so we can write the fact that represents the arrays on grid B as the quantity 5 times 3 times 2, to represent that there are 2 arrays. On grid A there are a total of 30 shaded squares. So grid A has a product of 30. And on grid B there are a total of 30 shaded squares, so the product of grid B is also 30.

Let’s go back to the first example from this section. The number sentence for grid A is, 5 times what that gives us 30? We found that this number sentence is, 5 times 6, because we have 6 items in each of our 5 groups. To write the number sentence for our second grid, we have 2 separate arrays that are different from each other, so we need to add these 2 arrays together. Each array has 5 groups. The first array has 4 items in each group, and the second array has 2 items in each group. So we read this number sentence as the quantity 5 times 4 plus the quantity 5 times 2 equals 30. If the products are the same for grid A and grid B then the factors can be written as equal to each other. 5 times 6 is equal to the quantity 5 times 4 plus the quantity 5 times 2. Remember that this shows that our 6 items in each group on grid A were broken up into 4 items in each group on the first array in grid B plus 2 items in each group on the second array in grid B.

Looking back at the second problem in this set of examples our first grid represents the fact 6 groups with 7 items in each group. Our number sentence is 6 times 7 equals 42. Our second grid represents 6 groups with 5 items in each group plus 6 groups with 2 items in each group which also equals 42. So our number sentence is written as the quantity 6 times 5 plus the quantity 6 times 2 equals 42. Our 2 number sentences are alike because the products are the same. Therefore if the products are the same then the factors can be written as equal to each other. 6 times 7 is equal to the quantity 6 times 5 plus the quantity 6 times 2. 5 items in each group on the first array in grid B plus 2 items in each group on the second array in grid B when added together equal the 7 items in each group on grid A.

Let’s look back at problem 3 from this set of examples. The factors are 2, 3, and 4. On grid A the 2 and the 3 have parenthesis around them, which means that we will multiply 2 times 3 first. We will draw a circle around the 2 times 3 to show that we are going to multiply these two numbers first. Arrow underneath our circle where we can put our product 2 times 3 equals 6. Since we haven’t multiplied by 4 yet, we will bring this down to create a new problem. 6 times 4 equals 24. So our product of the quantity 2 times 3 times 4 equals 24. Let’s look at the second fact represented by grid B. Again, the factors are 2, 3, and 4. This time the parenthesis, are around the 3 and the 4. So we will circle the 3 and 4 as these are the two numbers that we will multiply first, and place an arrow underneath our circled fact, so that we can put the product of this fact underneath. 3 times 4 is 12. Since we have not multiplied by 2 yet, we will bring this down in our problem. 2 times 12 is 24. So the product of 2 times the quantity 3 times 4 is 24. If the quantity 2 times 3 times 4 equals 6 times 4 which equals 24 and 2 times the quantity 3 times 4 equals 2 times 12 equals 24, then what can you conclude about these facts? We can conclude that 6 times 4 equals 2 times 12. We have the same factors in grid A as we did in grid B. What was different was which factors we multiplied together first. We showed in this example that it doesn’t matter what order we multiply these factors in. We will get the same answer each time.

Now let’s take a look back at problem 4 in this set of examples. The factors are 5, 2 and 3 for grid A. Since the 5 and 2 have parenthesis around them we will circle these factors to be multiplied first. 5 times 2 is 10, and bring down the third factor so that we can multiply the third factor by the product of the factors 5 and 2. 10 times 3 equals 30. On grid B we have the factors 5, 3 and 2. Since the factors 5 and 3 have parenthesis around them we will multiply these numbers first. 5 times 3 equals 15. We will bring down the third factor so that we can multiply it by the product of the factors 5 and 3. 15 times 2 equals 30. Looking at the product of grid A and the product of grid B, they are the same. Whether I multiplied the 5 and the 2 together first and then multiplied that product by 3 or I multiplied the the 5 and the 3 together first and then multiplied that product by 2, I got the same answer. So we have shown here that it doesn’t matter which two numbers I multiply first as long as I multiply all three of my numbers together. If 5 times 2 times 3 equals 30 and 5 times 3 times 2 equals 30, then what can you conclude about these facts? We can conclude that 5 times 2 times 3 is equals to 5 times 3 times 2.

Based on the examples we just completed we can use this graphic organizer later on to refer to when we need to multiply more than two numbers together. First we need to identify the problem that’s being asked. Next, we circle the factors to be multiplied first. We then multiply the two circled factors. Last, we multiply the product of the circled factors by the remaining factor.

Let’s apply the steps for multiplying to the next several problems. In example 1, the 6 and the 2 need to be multiplied first; 6 times 2 equals 12, and we bring down the 3 so that we can multiply it by the product of 6 times 2. 12 times 3 equals 36. So the product of this example is 36.

In our next example, we have the same factors, in the same order, but this time the parenthesis is around the 2 and the 3, so we will multiply these factors first. 2 times 3 equals 6, we bring down the 6 from the original problem, so that we can multiply it by the product of the factors 2 times 3. 6 times 6 equals 36. If the quantity 6 times 2 times 3 is equal to 6 times the quantity 2 times 3, then 36 is equal to 36. While the factors in these two examples are listed in the same order we did multiply them in the order that they are listed. However, we still ended up with the same answer for both problems. This shows that it doesn’t matter which two factors we multiply first, we still get the same answer.

In this next example, we have the factors 4, 9 and 2 that we are multiplying together. We will start by circling the two factors that we will multiply first. Since there are not any parenthesis around the factors in this problem, we will start at the left and work to the right. Our first two factors are 4 times 9, so we will circle these factors and find their product. 4 times 9 equals 36, we bring down the 2, because we have not multiplied the 2 yet, with the product of our first two factors. 36 times 2. Let’s look at the second example in this problem. We still have the factors of 4, 9 and 2 but they are in a different order this time. Again, since there are not any parenthesis around are factors we will work from left to right. So first, we will multiply 4 times 2. 4 times 2 equals 8. Now, let’s multiply the product of our first two factors with the third factor. In the first example we have 36 times 2 which gives us 72. And in our second example we have 8 times 9 which also gives us 72. So what can we conclude from these examples? If 36 times 2 equals 72, and 8 times 9 equals 72 then 36 times 2 is equal to 8 times 9. What we have proven in this set of examples is that it doesn’t matter what order the factors are in, in the problem, as long as we multiply all three of our factors together we will get the same product.