Divisibility by three and nine

Divisibility by three and nine

Pencil and paper
Overview
Using this Resource
Connecting to the Curriculum
Marking Student Responses
Working with Students
Further Resources
This task is about whether numbers are divisible by three or by nine.
Answer the following question without using a calculator.

 

a)
i)
 
 Is 25,396 exactly divisible by 3?                       YES /  NO     (circle one)
 
ii)
Explain how you could decide this without having to do the division.
 

 
 
 
 
 
 
 

 
 
b)
 
i)
 
Is 4,776 exactly divisible by 9?                         YES /  NO     (circle one)
 
ii)
Explain how you could decide this without having to do the division.
 

 
 
 
 
 
 

 
Task administration: 
This task is completed with pencil and paper only.
Level:
5
Curriculum info: 
Maths, Number and Algebra, Number Knowledge
Key Competencies: 
Using language, symbols, and texts
Keywords: 
basic facts, factors, divisibility
Description of task: 
Students identify and give rules for when numbers are divisible by three or by nine.
Curriculum Links: 
Key competencies
This resource involves explaining how they can tell is a large number is exactly divisible by 3 or 9. This relates to the Key Competency: Using language, symbols and text.
Learning Progression Frameworks
This resource can provide evidence of learning associated with
Multiplicative thinking, sets 8-9
within the Mathematics Learning Progressions Frameworks.
Read more about the Learning Progressions Frameworks.
Answers/responses: 
      Y10 (09/2008)
a)

 

 i)
ii)

 

 No
The sum of the digits is 25 which is not divisible by 3
or
Other correct explanations, including subtracting off amounts that are clearly divisible by 3.
Student examples:

  • Because the numbers at the end (396) divide exactly,
    [but] 25 doesn’t divide exactly.
  • Because 25 000 can’t be divided by 3000 without remainder.
  • 25 396 – 6 = 25 390;       25 390 – 390 = 25 000;
    25000 – 24000 = 1000;  1000 – 900 = 100; 100 – 99 = 1
 difficult
(both correct)
b)

 

 i)
ii)

 

 No
The sum of the digits is 24 which is not divisible by 9
or
Other correct explanations, including subtracting off amounts that are clearly divisible by 3.
Student examples:

  • 4776 – 36 = 4740;    4740 – 900 = 3840;
    3840– 1800 = 2040;  2040 – 1800 = 240;
    240 – 180 = 60;         60 – 54 = 6 not divisible
  • 4776 – 4500 = 276     276 – 270 = 6, not divisible
 very difficult
(both correct)

Based on a representative sample of 161 students.

Diagnostic and formative information: 
  Common error
a)

 

 Gives an incorrect rule based on the individual digits:

  • The number ends in a 3, 6, or 9 or the last digit is divisible by 3.
  • The last two or three digits are divisible by 3.
  • The first two or three digits are divisible by 3.

Gives an incorrect rule based on all the digits:

  • All digits are divisible by 3 (if they are, then the number will be divisible by 3, but if they are not then the number may still be divisible by 3 – i.e., sufficient but not necessary).).
  • Other incorrect rules about the sum of the digits.
b)

 

 Gives an incorrect rule based on the individual digits:

  • The number ends in a 9 or the last digit is divisible by 9.
  • The last two or three digits are divisible by 9.
  • The first digit(s) are divisible by 9.

Gives an incorrect rule based on all the digits:

  • All digits are divisible by 9 (if they are, then the number will be divisible by 9, but if they are not then the number may still be divisible by 9 – i.e., sufficient but not necessary).
  • Other incorrect rules about the sum of the digits.

These common errors may reflect that many students know that there is a relationship with the individual digits and divisibility by 3 or 9, but cannot accurately recall it.

Next steps: 
Gives an incorrect rule based on the individual digits:
The students need to know that rules for whether a number is divisible by 3 or by 9 depend on the sum of all the digits, and not on particular digits. The only cases where the divisibility test looks at only the last digit is when checking if a number is divisible by 2, 5, or 10.

Gives an incorrect rule based on all the digits:
Students often state either that:

  • a number is divisible by 3 if the sum of the digits is divisible by 3; or
  • a number is divisible by 9 if the sum of the digits is divisible by 9.

These statements are both correct, but they are misleading. This is an excellent opportunity to explore ideas of logic (such as necessary and sufficient conditions), and the role that counter-examples play.

  • The student is correct to say that “396 is divisible by 3 because all digits are divisible by 3.” This means it is sufficient for each digit to be divisible by 3.
  • However, the student is not correct if they say that 495 is not divisible by 3 because not all its digits are divisible by 3. Get them to check and see that 495 = 3 × 165. It is necessary for the sum digits of the number to be divisible by 3 for that number to be divisible by 3.

Developing a rationale for why the rules for divisibility work:
a)  Use decomposition/reduction strategies
Encourage students to see that if a number can be broken down into smaller parts that are each divisible by 3, then the whole number will also be divisible by 3. Start with easy examples.
   -   39 is divisible by 3 because 30 is divisible by 3 and so is 9.
   -   246 is divisible by 3 because 240 (or 24) is divisible by 3 and so is 6.
   -   1620 = 1500 + 120, and both of these are divisible by 3.
(see reference to Divisibility test, Book 8: Teaching Number Sense and Algebraic Thinking)
b) Look for patterns/rules in the sum of the individual digits
Get a substantial list of numbers; some which are divisible by 3, and some which are not.
Add the digits of all the numbers, and see what patterns emerge.

c) Explore why the sum of the digits rule works
See the box entitled “Student Exploration”.

Knows the rule for 3 but not the rule for 9:

  1. Ask the students: “Are all numbers that are divisible by 3 also divisible by 9?”
    Ask the students: “Are all numbers that are divisible by 9 also divisible by 3?
  2. Explore patterns in the sum of the digits of numbers divisible by 9 and those not divisible by 9.
  3. Explore strategies similar to the box entitled “Student Exploration”.

Extension: Multi-step use of the rule - (for students who knew the rules).
Get students exploring very large numbers to see if they are divisible by 3 or 9.
Students may need to "sum the digits of the sum of the digits" i.e. repeat the process iteratively. Student examples from part a) of the question:
            2 + 5 + 3 + 9 + 6 = 25;     2 + 5 = 7;   7 is not divisible by 3;   so 25396 is not divisible by 3.
Ask:
If the following number is divisible by 9 and to justify their answer.
                        19 823 764 554 673 829 182 737
Answer:
Yes,because he sum of the digits is 117 which is divisible by 9, as 1 + 1 + 7 = 9, which is divisible by 9.

Student exploration
Get students to explore the conjecture that numbers are divisible by 3 if the sum of their digits is divisible by 3.

  • Can they find an example where it doesn’t work?
  • Can they find an explanation why it works?
  • Can they show that it always work?

This is the hardest part, but leads to algebraic understanding.

The key is in modular arithmetic, i.e. the remainder when a number is divided by 3. Lay out the numbers as follows

1
4
7
10
97
100
997
1000		 2
5
8
11
98
101
998
1001		 3
6
9
12
99
102
999
1002
  • Numbers in the last column are divisible by 3.
  • Numbers in the first two columns have a remainder of 1 and 2 respectively.
  • The remainder of the division by 3 is called mod 3.
  • So 2 (mod 3) = 5 (mod 3) = … = 98 (mod 3) = 101 (mod 3) = … = 2

The other key factor is that our number system is base-10.

  • Because 3 × 3 = 9, then 10 has a remainder of 1 when it is divided by 3. But so does 100, 1000, 10 000 etc.
  • So 1 (mod 3) = 10 (mod 3) = 100 (mod 3) = 1000 (mod 3) = … = 1
  • So for a three digit number, 495 say we can do the following:
    495 = 4 × 100 + 9 × 10 + 5 × 1 (place value decomposition)
  • But because 1000, 100, 10, 1 all have a remainder of 1 when divided by 3, then 495 (mod 3) = [4 × 100 + 9 × 10 + 5 × 1] (mod 3) = 4 × 1 + 9 × 1 + 5 × 1 = 4 + 9 + 5.
  • So if 4 + 9 + 5 is exactly divisible by 3 so is 495.
For further information about divisibility tests, refer to the following: