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Learning Math Home
Number and Operations Session 5, Part B: Divisibility Tests
 
Session5 Part A Part B Part C Homework
 
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Session 5 Materials:
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Session 5, Part B:
Divisibility Tests

In This Part: Developing Testing Rules | Divisibility Tests for 2, 5, and 10
Divisibility Tests for 3 and 9 | Divisibility Tests for 4 and 8 | Divisibility Test for 11

Knowing how divisibility tests work helps us think carefully about factors and multiples. All the tests for divisibility so far are for factors of 10. What about other numbers, like 3 or 9? Let's start with divisibility by 9, because 9 is one less than 10.

First, let's look at the numbers that are divisible by 3 and/or by 9:

Row

Numbers

1

1

11

21

31

151

2461

2

2

12

22

32

152

2462

3

3

13

23

33

153

2463

4

4

14

24

34

154

2464

5

5

15

25

35

155

2465

6

6

16

26

36

156

2466

7

7

17

27

37

157

2467

8

8

18

28

38

158

2468

9

9

19

29

39

159

2469

10

10

20

30

40

160

2470

Blue: Divisible by 3, but not 9
Red: Divisible by 9 and 3

Notice the sum of the digits of each number in red. If the sum of the digits of a number is divisible by 9, then the number is divisible by 9. Why does this work?

Consider the four-digit number 1,116, which in expanded form is (1 • 1,000) + (1 • 100) + (1 • 10) + (6 • 1). Below, the number is modeled with base ten blocks:

The sum 1,000 + 100 + 10 + 6 could be expressed as (999 + 1) + (99 + 1) + (9 + 1) + 6, which can be re-expressed as (999 + 99 + 9) + (1 + 1 + 1 + 6) or 9(111 + 11 + 1) + 9. This shows that the number 1,116 is divisible by 9, because it can be expressed as 9 more than a multiple of 9, which in itself is a multiple of 9.

The distributive law also allows us to pull a 9 out of each piece of the sum 999 + 99 + 9 + 9 = 9 • (111 + 11 + 1 + 1), thus showing that the sum is divisible by 9.

Let's apply a similar analysis to the number 261:

261 =
(2 • 100) + (6 • 10) + (1 • 1) =
(2 • [99 + 1]) + (6 • [9 + 1]) + (1 • 1) =
([2 • 99] + [6 • 9]) + ([2 • 1] + [6 • 1] + [1 • 1]) =
9 • ([2 • 11] + 6) + 9

Thus, we've shown that the above sum is divisible by 9, and, as a result, that the number 261 is divisible by 9.

Let's test another number, for example, 3,455:

3,455 =
(3 • 1,000) + (4 • 100) + (5 • 10) + (5 • 1) =
(3 • [999 + 1]) + (4 • [99 + 1]) + (5 • [9 + 1]) + (5 • 1) =
([3 • 999] + [4 • 99] + [5 • 9]) + ([3 • 1] + [4 • 1] + [5 • 1] + [5 • 1])

Factoring out 9 from the first parenthesis, we get:

3,455 = 9 • (333 + 44 + 5) + (3 + 4 + 5 + 5)

Since we know that 9 • (333 + 44 + 5) is divisible by 9, we only need to examine the second parenthesis. Note that this is the same as the sum of the digits of the original number! The sum is 17, which is not divisible by 9. Thus, 3,455 is not divisible by 9.

Because 3 is a factor of 9, the divisibility test for 3 is related to the test for 9. If the sum of the digits of a number is divisible by 3, then the number is divisible by 3.

The number 3,455 is not divisible by 3, since 3 + 4 + 5 + 5 = 17, which is not divisible by 3. Why does this work?

Applying the same analysis we've been using, let's determine why 3,455 is not divisible by 3:

3,455 =
(3 • 1,000) + (4 • 100) + (5 • 10) + (5 • 1) =
(3 • [999 + 1]) + (4 • [99 + 1]) + (5 • [9 + 1]) + (5 • 1)=
([3 • 999] + [4 • 99] + [5 • 9]) + ([3 • 1] + [4 • 1] + [5 • 1] + [5 • 1])

Factoring out 9 from the first parenthesis, we get:


 

Problem B3

Solution  

Since you know the tests for 2 and 3, can you devise a divisibility test for 6? Use the following chart to check your rule:

Row

Numbers

1

1

11

21

31

151

2461

2

2

12

22

32

152

2462

3

3

13

23

33

153

2463

4

4

14

24

34

154

2464

5

5

15

25

35

155

2465

6

6

16

26

36

156

2466

7

7

17

27

37

157

2467

8

8

18

28

38

158

2468

9

9

19

29

39

159

2469

10

10

20

30

40

160

2470


 

Problem B4

Solution  

What do you notice about the factors you used to check for divisibility by 6? What is the divisibility test for 15?



video thumbnail
 

Video Segment
In this video segment, Vicky and Nancy explore different properties of numbers that are multiples of 6 in order to figure out the rule for divisibility by 6. Watch this segment after you've completed Problems B3 and B4.

Did you use similar methods to come up with a rule?

If you are using a VCR, you can find this segment on the session video approximately 7 minutes and 17 seconds after the Annenberg Media logo.

 


video thumbnail
 

Video Segment
In this video segment, Donna explains her group's thinking in finding the divisibility test for 6. With Professor Findell, she further explores the relationship between the numbers 2 and 3, which together guarantee divisibility by 6.

If you are using a VCR, you can find this segment on the session video approximately 12 minutes and 1 second after the Annenberg Media logo.

 

Next > Part B (Continued): Divisibility Tests for 4 and 8

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