AO elaboration and other teaching resources
This unit looks at simple rules that will decide when any number has the factors 2, 3, 4, 5, 6, 8, 9, 10, 11, and 12. It also looks at how we can quickly tell what remainder a number has when it is divided by 3 or 9. We also give the proofs of these rules. These ideas are tested by word problems.
- explain divisibility rules
- find whether a number is a multiple of 2, 3, 4, 5, 6, 8, 9, 10, 11, or 12
From time to time it is useful to know how to test a number to see if it has a given small factor. In this unit we look at such tests for 2, 3, 4, 5, 6, 8, 9, 10, 11, and 12.
You will notice that 7 is omitted from these tests. This is not because there is not a test for 7. Actually there are several. However, these tests are not simple and so we have decided not to include them. One such test can be found on www.primepuzzles.net/puzzles/puzz_101.htm. This actually has a lot of other tests too that you might like to think about. If you do a search on ‘divisibility tests’ or ‘divisibility rules’ you will find several other sites which cover similar ground.
Probably, given that most students these days have calculators, these tests are not as useful as they once were, though knowing the factors of a given number is still important because of applications in cryptography. And what’s more they will all be getting a chance to practice quite a bit of arithmetic.
Perhaps the main reason for doing these tests is that it shows the power of algebra. At this stage in their careers, your students have probably seen algebra mainly for solving equations. In this unit we give a completely different use. Here we use algebra to prove things, specifically we’ll prove that the tests really work.
Let’s think about the ‘3 rule’. We could prove this for 2-digit numbers by showing that it works for all ninety 2-digit numbers. But the proof that works for a 2-digit number can be turned into algebra and then we can see that it works for ALL 2-digit numbers AT THE ONE TIME. This is the real power of algebra! One calculation with letters a and b, can replace 90 calculations with specific numbers.
But then we can do the same thing for 3-digit numbers. Nine-hundred specific calculations can be replaced by one algebraic manipulation. This process can be repeated for 4-digit numbers, 5-digit numbers, and so on. However, using more sophisticated algebra (mainly the idea of summations with subscripts), the ‘3-rule’ can be proved in a few lines for a number with ANY number of digits! So a few lines of algebra can do what might otherwise require an infinite number of calculations!!
We have given this general proof in the Teachers’ Notes to session 2. We don’t expect all but your most able students to be able to understand it. But you may like to see it for yourself and you will have a student from time to time who will lap it up.
Why do all these rules work? It’s simply because we use the decimal system. The Romans couldn’t use these rules because they didn’t have the number system to base them on. Neither could they hope for any rules at all that would simplify divisibility. Ask students to investigate the fundamental role that the decimal system plays in this exploration of patterns.
These rules are things that students could be reminded of from time to time through the rest of the year.
Once again here the philosophy that we use through the unit is that of ‘guided discovery’. As much as possible we try to set up situations that give students the opportunity to find things out for themselves. In doing so they are essentially doing research. This can be a lot more interesting and lead to better learning than just providing students with the results and then letting them apply those results.
In this session we refresh students’ knowledge of the rules that can be used to decide when a number is divisible by 2 or 5 or 10. We then look at divisibility by 3.
Students should know the meaning of the following words: divisible, factor, multiple, even, odd, palindrome, remainder.
We have used the phrase ‘if and only if’ here. This is an important condition in mathematics as it tells us that two things are mathematically the same, that they are mathematically equivalent. If you think that it is too hard for most of your students then you may want to just stick with ‘if’. However, it may well be worth explaining the term for your more able students.
So when we say that a number n is divisible by 2 if and only if the last digit of n is 0, 2, 4, 6, or 8, we mean two things. First, that n is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. Second, if n is divisible by 2, then the last digit of n is 0, 2, 4, 6, or 8.
Logically then, being divisible by 2 and ending in 0, 2, 4, 6, or 8, are equivalent.
Start this session with the whole class working together. Ask them the following questions and try to get the responses in brackets.
What can you tell me about these numbers? 8, 20, 10, 16, 22, 6. (These are all even numbers.)
What can you tell me about even numbers? (2 goes into them. They all end in an even number.)
Why would it be useful to know if we had an even number of sweets? (They can be shared equally between two people.)
Give me some 3-digit even numbers.
Give me some 5-digit even numbers.
I want to make an even number with 4 digits that starts 385. What could the fourth digit be? How many answers could there be? (0, 2, 4, 6, 8.)
I have a 4-digit even number 791d. What could d be?
I have a 4-digit even number 307d. What could d not be?
Give me some 3-digit numbers that are not divisible by 2?
What rule can I use to decide whether or not a number is divisible by 2 or not? (A number is divisible by 2 if and only if its last digit is 0, 2, 4, 6, or 8.)
Display this rule on the classroom wall and/or get the students to write it in their note books.
Repeat the last sequence of questions with numbers that are multiples of 10. Complete this part with the rule: a number is divisible by 10 if and only if its last digit is 0.
Repeat the sequence of questions again, this time with numbers that have a factor of 5. Complete this part with the rule: a number is divisible by 5 if and only if its last digit is 0, or 5.
Let the class work in groups on the problems on Copymaster 1.
Then turn the class’s attention to divisibility by 3 by letting them work on Copymaster 2.
Bring the whole class together to discuss the various groups’ conjectures on what numbers are divisible by 3. Do not settle on any particular rule as being correct or not. Accept what the students say but insist that they justify what they have discovered BY USING EXAMPLES ONLY. (We will look at a proof in the next session.)
In this session we look at proving the rules for divisibility by 2, 5 and 10. We also establish and prove the rule for divisibility by 3.
Divisibility by 2: Numbers less than 10 clearly satisfy the rule. Any number N greater than 10 can be written as 10a + b, where 0 ≤ b ≤ 9. So N = 10a + b. But 10a is even. So if b is even then N has to be, while if N is even b has to be. So we have the rule that N is even if and only if b is even. Since b is a digit, b has to be 0, 2, 4, 6, 8. So N is even if and only if its units digit is one of 0, 2, 4, 6, or 8.
This may not be the only way to do this.
Divisibility by 10: Any number N greater than 10 can be written as 10a + b, where 0 ≤ b ≤ 9. So N = 10a + b. But 10a is divisible by 10. So if b is divisible by 10 then N has to be, while if N is divisible by 10, b has to be. But the only digit that is divisible by 10 is 0. So we have the rule that N is divisible by 10 if and only if b is 0. Integers that are divisible by 10 have to end in zero.
Divisibility by 5: Numbers less than 10 clearly satisfy the rule. Any number N greater than 10 can be written as 10a + b, where 0 ≤ b ≤ 9. So N = 10a + b. But 10a is divisible by 5. So if b is divisible by 5 then N has to be, while if N is divisible by 5, b has to be. The only digits that are divisible by 5 are 0 and 5. So we have the rule that N is divisible by 5 if and only if b is 0 or 5.
Divisibility by 3: Let’s look at a 4-digit number. Let N = abcd. Then N = a x 1000 + b x 100 + c x 10 + d = a x (999 + 1) + b x (99 + 1) + c x (9 + 1) + d = 999a + 99b + 9c + (a + b + c + d ) = 3 x (333a + 33b + 3c) + (a + b + c + d) = 3R + (a + b + c + d). So N = 3R + (a + b + c + d). So if (a + b + c + d) is divisible by 3 then so is N and if N is divisible by 3 so is (a + b + c + d).
Once again we get N is a multiple of 3 plus the sum of the digits of N and the argument from the first paragraph follows.
Note that 9…9 and 3…3 means that you have to use i nines and i threes.
This subscript notation is hard to follow at first but can be quite useful. We suggest that you show it only to very able students. You will need to explain to them that the sigma is the Greek letter S and that S stands for Sum. So the notation is about summing things. It works like this:
What happens is that you start with putting i equal to 1; then you add what happens when i = 2; then add what happens when i = 3; then add what happens when i = 4; then add what happens when i = 5. Because the number at the top of the sigma is 5, we stop with i = 5.
In the same way we get:
In this session we want to answer the ‘why’ question. Mathematicians always want to know ‘why?’.
Why does the rule for division by 2 really work?
Why does the rule for division by 10 really work?
Why does the rule for division by 5 really work?
So let’s look at the first one first.
Why does the rule for division by 2 really work?
What reason can you give for this?
Discuss their ideas and help them develop their own proof or lead them to see the reason given in Teachers’ Notes. Let them be in the driving seat as much as possible.
Let them go into their groups to consider the cases of 10 and 5, in that order. When they have worked on that for a while bring them back together to discuss what they have found. Get them the opportunity to make appropriate notes in their notebooks.
Now move on to the problem that we were working on at the end of the last session.
So what about 3 then?
From what we did in the last session, what rules do you think hold for divisibility by 3?
Discuss their ideas and try to settle on a correct rule. Discuss how you might prove the rule.
Let them work in their groups to prove the ‘3 rule’. Tell them that it might help if they can prove it for a lot of numbers but that it would be good to be able to do it, say, for all 2-digit numbers.
Bring the class together to consider what they have done. Let them present their ideas in front of the class. Get a good proof constructed on the board. Display this rule on the classroom wall.
(The proof we have given first in Teachers’ Notes is reliant on a particular number of digits in the number N. The more able students might like to think about how they could prove the rule for ANY number of digits – see Teachers’ Notes.)
Let them write up the rule’s proof fro a specific number of digits in their notebooks.
Give the students a chance to get to grips with these rules by working on the questions in Copymaster 3.
In this session we give the students the opportunity to discover rules for divisibility by 4, 6, 7, 8, 9.
Divisibility by 4: Take the last two digits of the number. If they, as a 2-digit number are divisible by 4, then the number is and vice versa. Any number N can be written as 100a + bc, where 0 ≤ bc ≤ 99. But 100a is divisible by 4 since 4 goes exactly into 100. So if bc is divisible by 4 then N has to be, while if N is divisible by 4, then bc has to be.
Divisibility by 6: Let N = 6k be a multiple of 6. Then N = 2 x 3 x k. So if both 2 and 3 divide N, then so does 6. So the rule for 6 is the rules for 2 along with the rule for 3.
Divisibility by 7: There are several rules for 7 but they are not much better than dividing by 7 directly.
Divisibility by 8: This is similar to divisibility by 4. Let N = 1000a + bcd. Then N is divisible by 8 if and only if the 3-digit number bcd is divisible by 8.
Divisibility by 9: This is similar to divisibility by 3. Let’s look at a 4-digit number. Let N = abcd. Then N = a x 1000 + b x 100 + c x 10 + d = a x (999 + 1) + b x (99 + 1) + c x (9 + 1) + d = 999a + 99b + 9c + (a + b + c + d ) = 9 x (111a + 11b + c) + (a + b + c + d) = 9R + (a + b + c + d). So N = 9R + (a + b + c + d). So if (a + b + c + d) is divisible by 9 then so is N and if N is divisible by 9 so is (a + b + c + d).
Revise the rules that you have been investigating over the last two sessions. This can be done in a number of ways. For instance, you could read out a sequence of numbers and ask the students to write down what factors they have. You could also read out a sequence of numbers and get students to say aloud what the factors are. You might also let them suggest some numbers too. It would help if numbers had more than one of 2, 3, 5, and 10 as factors.
Remind the class that for numbers under 10, we haven’t yet looked at 4, 6, 7, 8, and 9. Let them investigate these numbers in their groups to see if they can find any nice rules for divisibility. Circulate around the groups and support as needed. Allow groups time and support to discover any given rule.
Get the class together at a suitable moment and discuss what they have found. Give them time to write the rules in their notebooks. Display the rules on the classroom wall.
Let them use their new-found knowledge on Copymaster 4.
This session it is the turn of 11 and 12. If there is time, you might also like to look at how to test for remainders after dividing by 3 and 9.
Divisibility by 11: This is a twist on the divisibility by 3 (or 9) rule. We do it for a number with several digits. Let N = abcdefghjk. Let O = k + h + f + d + b and let E = j + g = e + c + a. Then N is divisible by 11 if and only if O – E is divisible by 11. (This can be proved in a similar way to the ‘3 rule’. However, you need to realise that 10 = 11- 1; 1000 = 1001 – 1; 100000 = 100001; etc. and 100 = 99 + 1; 10000 = 9999 + 1; etc.).
Divisibility by 12: This is similar to divisibility by 6. If N = 12k, then N = 3 x 4 x k. So to see if N is divisible by 12 or not we only have to check for divisibility by 3 and 4.
Revise the rules that we have been working on.
It turns out that there are reasonable rules for divisibility by 11 and 12.
Can you think what the rule for 12 might be?
Discuss this with the class.
The ‘11 rule’ may be difficult for students to find on their own so we suggest that you either tell them what it is or lead them carefully to the right answer.
Let the class do the problems on Copymaster 5 in their groups.
Give the students time to write up the rules for 11 and 12 in their notebooks. Display this rule on the classroom wall.
Then let them work in their groups on Copymaster 6.
Bring the class together to discuss the remainder situation. How can you tell what the remainder is when you divide by 3? Let them write up a summary in their notebooks. Display this rule on the classroom wall.
If there is time, let them work in their groups on Copymaster 7.
Bring the class together to discuss the remainder situation. How can you tell what the remainder is when you divide by 9? Let them write up a summary in their notebooks. Display this rule on the classroom wall.
In this final session we revise the rules of the previous sections of the unit.
The Number Property game: This is a game that can be played with not just numbers but with almost anything. Let’s talk about the version that we want to play in this unit and then suggest other ways that it could be used.
Here we want to get the students thinking about the divisibility properties that we have developed in this unit. So the game is played by you (or one of the students) thinking of a particular divisor. Tell them that you are going to give a sequence of numbers that all have a common property. When anyone thinks they know what the property is they should put up their hand and then give another number with that property. THEY ARE NOT ALLOWED TO SAY WHAT THE PROPERTY IS! If the number has the right property then you will say ‘yes’; if it doesn’t you will say ‘no’. Then you will say another number with the same property and then expect someone else to say a number with that property. Keep going until most of the class has contributed a number.
This game can be played with any patterns at all. These may not necessarily be number patterns. For instance, you might use European countries or famous artists or writers. It’s a useful way to focus students’ attention on an idea that you may want to revise or develop.
Start this session with the whole class together. Play the ‘Number Property’ game with first you as the person deciding on the property and then one of the class taking the leading role. (You may want to give them time to produce a list of numbers that have the required property. It isn’t easy making up numbers on the spot.)
Get the students to play the ‘Number Property’ game in their groups.
In their groups the students can also write 10 questions to ask the other groups. These questions should relate to the rules that we have been discussing here. Encourage them to produce one question that needs a proof. (Some of these questions can be modelled on the questions in the Copymasters.) Each group must produce answers to each of their questions.
Each group takes turns in ‘administering’ their ‘test’. Marks could be given for correct answers. You could also get them to give bonus marks for original questions.