Sticking Together

The aim of this activity is to explore the strategy of partitioning a factor in a multiplication problem to make it easier to solve. This is most often done using place value to break the number into smaller chunks.
To solve 4 x 132 using this strategy, 132 is partitioned into 100 + 30 + 2, each of these parts is multiplied by 4, and the products added together: 4 x 132 = (4 x 100) + (4 x 30) + (4 x 2)
= 400 + 120 + 8
= 528
See the notes for Bean Counters (page 21) for a rationale for students learning and using a broad range of multiplicative strategies and for links to the NDP strategy teaching model.
Encourage the students to manipulate place value materials to solve the problems in questions 1 and 2. Beans and canisters or tens money ($1 coins and $10 notes) could be used in place of bundles of sticks. Watch that the students are seeing the bundle of 10 as a unit rather than treating them as 10 single sticks.
Challenge students who simply skip-count the bundles in the picture (10, 20, 30, …) to come up with an alternative way of finding out how many sticks there are in the bundles, using their multiplication facts as Paul does.
Some students may not have made the connection between their basic facts and how they can be applied to multiples of 10 or 100. Emphasise this link by saying “2 tens” rather than “twenty” when talking about how knowing 4 x 2 = 8 might help Paul with working out 4 x 2 tens. “Twenty” does mean “2 tens”, but this is not necessarily obvious. See the NDP links at the end of the notes for this activity for learning experiences that help students to decode English number words such as
“-teen” and “-ty” (which both mean “ten”).
To promote generalisations, ask questions such as How does knowing 5 x 7 = 35 help you to work out 5 x 70 or 50 x 7 or 50 x 70? Have the students check their predictions on a calculator and talk about patterns they can see. Ask them to try to generalise a rule.
Promote imaging for questions 3–4 by asking the students to describe what they would see and do with the sticks to solve 5 x 32 using Paul’s method. Expect responses such as: “I would have 5 groups, and each group would have 3 bundles and 2 single sticks. I’d find out how many sticks there are in the bundles by working out 5 groups of 3 tens, which is 15 tens or 150. Then I’d find out how many single sticks there are by working out 5 groups of 2, which is 10. So the total number of sticks in the bundles and singles is 150 + 10, which is 160.” Go back to using materials if needed.
Use question 5 as a formative assessment opportunity to see if your students are able to use number properties to solve the problem using place value partitioning. Go back to imaging if needed by asking the students to describe what the materials would look like and what they would do with them to solve the problem. 100 sticks can be shown by putting 10 bundles of 10 in a ziplock bag or an ice cream container.
Use question 6 as a lead-in to the students writing their own problems in question 7. Talk about the sorts of numbers that Paul’s strategy is useful for, encouraging the students to see this strategy as one tool in their toolbox of strategies to be used as needed.
After the students have worked on question 6, ask: Even though 898 is a smaller number than 2 153, it was more complicated to solve 4 x 898 using this strategy than it was to solve 4 x 2 153. Why was this? (For 4 x 898, there were several numbers that needed renaming when adding, so it was harder to keep track of the mental calculations.)
Look back at the previous problems in the activity and record them in a list: 4 x 23, 3 x 42, 5 x 32, 8 x 21, 3 x 152. Ask What is similar about all of these problems? (One of the numbers is a single digit, and the other uses only small digits.)
Extension
See the strategy toolbox reference display idea outlined in the notes for “Bean Counters” (page 22).
 

Answers to Activity

1. a. 4 x 2 = 8, so 4 x 2 tens = 8 tens, or 80.
b. 4 x 3 = 12 because there are 4 groups of 3 single sticks.
c. 92. ([4 x 20] + [4 x 3] = 80 + 12 = 92. So 4 x 23 = 92.)
2. a. 3 groups of 42 or 3 x 42 = 
b. 120. (3 x 4 tens = 12 tens, which is 120. Or, using Paul’s hint, 10 x 10 = 100 and
2 x 10 = 20. 100 + 20 = 120)
c. 6. (3 x 2 = 6)
d. 126 sticks. 120 + 6 = 126, so 3 x 42 = 126.
3. a. A possible picture is:

b. 150. (5 x 3 = 15, so 5 x 3 tens = 15 tens, which is 150.)
c. 10. (5 groups of 2 = 10)
d. 160. (150 sticks in the bundles + 10 single sticks = 160, so 5 x 32 = 160.)
4. Paul could use his sticks like this:

Then he could go: 8 x 2 = 16, so 8 x 2 tens = 16 tens or 160. (Or he could go: 8 x 2 tens = 8 x 20 = 160.) 8 x 1 = 8. 160 + 8 = 168. So 8 x 21 = 168.
5. The 1 in 152 represents 1 hundred, the 5 represents 5 tens, and the 2 represents 2 ones. Paul could think of the digits as being “hundreds”, “tens”, and “ones”.
3 x 1 hundred = 300
3 x 5 tens = 15 tens, which is 150
3 x 2 ones = 6 ones
300 + 150 + 6 = 456. So 3 x 152 = 456.
6. a. i. (4 x 2 000) + (4 x 100) + (4 x 50) + (4 x 3) = 8 000 + 400 + 200 + 12 = 8 612
ii. (4 x 800) + (4 x 90) + (4 x 8) = 3 200 + 360 + 32 = 3 592
b. Paul’s strategy was easiest to use with the first problem. The second problem is harder to keep track of in your head because there are lots of numbers that need renaming.
c. 898 is close to a tidy number, so a useful strategy would be to use the tidy number and then compensate: (4 x 900) – (4 x 2) = 3 600 – 8 = 3 592
7. Problems will vary. Your problem might have one number that has 1 digit, and the other number might have 2, 3, or more digits. If you use small digits, the strategy will be easier to use. Examples could be:
3 x 34 = (3 x 30) + (3 x 4) = 90 + 12 = 102, or
5 x 162 = (5 x 100) + (5 x 60) + (5 x 2) = 500 + 300 + 10 = 810. Use a calculator to check that your answer is correct before you give your problem to your classmate.