Understanding Divisibility Rules
Brian D. Rude, 2011
Divisibility rules run from the very simple, to the not at all simple. Divisibility by 7 is a hard one, but it can be understood with a little bit of effort and time.
I'm not sure how important divisibility rules ought to be in the teaching of arithmetic. They are certainly good to know, but that does not mean that they should be considered an important foundation for algebra, or anything else. Is it worth the time it takes to learn the divisibility rules in elementary school? I don't know. That would depend on how much time is available, and what the competition for that time is.
But if there is time to teach the divisibility rules, should they be taught by rote, or can we teach them with understanding? Certainly in the teaching of math it is always desirable to teach for understanding as much as we can. But again we must make trade offs. Teaching something with understanding takes time. Is the time available? That depends on a lot of things, including what the competition is.
I think understanding of the divisibility rules is possible in arithmetic in elementary school, before algebra. But understanding would certainly seem to be much easier after the student has a year or so of algebra. At that time the language and notation conventions of algebra can be used. To teach understanding of the divisibility rules requires some sort of notational conventions. I think standard algebraic notation conventions ought to be used, but since, in arithmetic, those conventions, or at least some of them, must be taught along with the ideas being considered, the time demands are not trivial. For the moment I simply want to explain the divisibility rules in a way in which I think arithmetic students can makes sense of them. I make no assertions on what ought to be done, only some ideas on what might be done.
Let us start with the rule for divisibility of two, which is very simple. You look at the last digit in the number, the digit in one's place. If it is divisible by two, then the entire number is divisible by two.
Example 1: Is the number 13,085 divisible by two?
Answer: No. You only need to look at the digit in one's place, the five. Since it is not divisible by two, then the entire number, 13,085 is not divisible by two.
I think this is probably easily understandable by students before they have any algebra. Indeed when whole number division is pretty well established it is quite understandable. The students have been dividing by 2, and are aware, at least intuitively, that the remainder will be either 0 or 1. But to prepare the way for understanding other divisibility rules, we should look closely at this divisibility rule. When asked to explain this rule, or an example that applies this rule, I presume a child would say something like the following. "If you take away the five, you're left with 13,080, and obviously that's divisible by two. So therefore . . . . . . it's obvious."
Well, it may seem obvious, but we would profit by investigating. Can we set up a few rules for divisibility that are understandable to elementary school children, and that will make all the divisibility rules comprehensible? I suggest the following: I have one definition and two rules to establish. After these are established, then every thing I talk about in divisibility can be mathematically justified as a logical consequence of these three principles.
The first of the three principles is simply the definition of divisibility. I will state it this way:
If a number contains x, with nothing left over, then it is divisible by x.
Logically or mathematically this is not a very good definition. It is just one way of stating what we mean by divisibility. But remember that we want to be pedagogical, at the moment. We need a statement of what we mean by divisibility that is meaningful to kids. My thinking is that the phrase, "contains x, with nothing left over" would make sense to kids and be phrasing that they could remember. Some one who actually teaches math to elementary school kids might well come up with a better phrasing yet. I have called this first principle the definition of divisibility, but I'm not sure that's the best way to present it to kids. I think it might be better to call it a rule or a principle. I will call it the "basic principle". I think kids can probably understand that it is a definition, but I think calling it the "basic principle" might give kids a better handle on it.
The next two principles of divisibility are rules. Again we want to formulate them in ways that are easily understandable by children. It is desirable for them to be mathematically precise to whatever degree is possible, but making sense to children is the first consideration..
Rule number one: If at least one of two factors of a number is divisible by x, then the whole number is divisible by x.
Rule number two: If both addends of a number are divisible by x, then the whole number is divisible by x.
Are the terms "factors" and "addends" meaningful to elementary age children? I think they should be. If they are not meaningful before studying divisibility it would seem worth while to explain them and expect children to use the terms correctly. Even if the children are used to these two terms, know that factors refer to multiplication and addends refer to addition, some explanation and examples should be given.
Instead of saying "at least one factor" in rule one, would it be better top simply say "one factor"? I am inclined to think "at least one factor" is the better way to say it. We could argue that "at least one factor" and "one factor" are logically equivalent in this context, so we should use the more economical expression. But again strict logic is not the most important thing.
We will restrict ourselves to whole numbers at this point, of course, not fractions and decimals. Divisibility, as we are using the term here, is defined only for whole numbers. Hopefully this is obvious to the teacher, and perhaps obvious to the children. But that does not mean that it doesn't need to be explicitly pointed out.
Example of rule number 1: 36 can be decomposed into the factors 18 and 2. 18 is divisible by 3. Therefore 36 is divisible by 3. It doesn't matter that 2 is not divisible by 3. It doesn't have to be. Rule 1 works if only one of the factors is divisible.
Another example of rule number 1: 135 can be decomposed into the factors 15 and 9. 15 is divisible by 5. Therefore 135 is also divisible by 5.
Example of rule number 2: 84 can be decomposed into the addends 70 and 14. Both 70 and 14 are divisible by 7. Therefore their sum, 84 is divisible by 7.
Another example of rule number 2: 78 can be decomposed into the addends 60 and 18. Both 60 and 18 are divisible by three. Therefore 78 is divisible by 3.
Should we call these rules theorems? That would imply they are derived from some other rules, which is probably not the case at the moment. It might be argued that both rules derive logically from the arithmetic that has been developed, but we don't necessarily want to get deep into just how that works at this point, Can we call these rules axioms? I think we can. They make sense so we accept them. Is it too early to introduce the terms "axiom" and "theorem"? I don't know. The distinction is that an axiom is accepted without proof, as just a starting point. A theorem is not accepted without proof. We don't call a rule a theorem unless we can show that it is a logical consequence of other facts or rules that we have shown to be true. My guess is that it would be entirely appropriate to introduce these terms in this context. But I have no experience to base this on. I haven't taught arithmetic recently.
I would think that at this point students would have to do a lot of simple problems applying these two rules. The goal, at least the goal I have in mind, is to make these two rules concrete enough so that they make sense to students. I presume this can only be done operationally, not definitionally. That means doing problems. That means homework. (Of course the term "homework" doesn't necessarily mean it is to be done outside of classroom hours. It means, to me at least, that problems be done individually in an open time frame. There may be plenty of time in school hours for this to be done in school.)
Would it make sense to turn these rules around? Consider this example: 56 can be decomposed into the two addends 35 and 21. Both 35 and 21 are divisible by 7. Therefore by rule 2 it must be that 56 is divisible by 7, and of course we know that it is. But 56 can be decomposed differently. It can be decomposed into 40 and 16, neither of which is divisible by 7. Does rule 2 therefore tell us that 56 is not divisible by 7? In the language of logic we would say that the inverse of a rule may or may not be true. The inverse of rule two is definitely not true. Would it make sense to try to define inverse at this level and talk about it? I would expect that it would not. But perhaps it might be useful to bring up an example like this and to point out that rule 2, or any other rule, means exactly what it says, and no more. If we decompose 56 into 40 and 16 then rule 2 says nothing. It doesnít apply. Would we only cause confusion by bringing up an example such as this? Or would it serve to emphasize that rules should be read and applied precisely? I donít know.
I will now take the divisibility rules one by one and try to show how the above three principles, one definition and two axioms, and deductive reasoning, will explain everything. By "explain everything", of course I am talking about explaining to elementary school students. Our goal is not to find the precise wording that would be required in a mathematical publication. Our goal is to be comprehensible to students, to make everything seem like nothing more than common sense to them. We will first return to the rule for divisibility by 2.
The divisibility rule for 2 is simply to look at the last digit in the number. If it is even, that is 0, 2, 4, 6, or 8, then the entire number is divisible by two. Remember that this may be obvious to students, and therefore a little hard to explain why we might want to apply rules and deductive reasoning. This may not become clear to them until much later, after we have dealt with the divisibility rules of other numbers.
So let us take the example of 15,484. Why do we know it is divisible by 2?
By the first principle, the "basic principle", which I have called a definition, we can say that 15,484 can be decomposed into 7742 and 2. Therefore 15,484 "contains" 2. Problem solved! Proof complete.
However the whole idea of having divisibility rules is to have shortcuts. We didn't use a shortcut in the above analysis. We simple used the "basic principle", the definition of divisibility. We divide by 2 in order to show that 15,484 can be decomposed into 7742 and 2. There is no remainder left over. The shortcut is that we only need to look at the last digit, 4 in this problem, to see that the whole number, 15,484 can be divided by two and there will be no remainder. How does this work? Why will the rule always work?
15,484 can be decomposed into the sum of 15480 and 4. By rule two both of these addends must be divisible by 2 if we are to say that the whole number is divisible by 2. The 4 is obvious, but what about the 15480? Indeed it probably is obvious to children that 15480 is divisible by 2, but how can we relate it to the three principles?
It seems that at this point a theorem is appropriate. The theorem would be that any number ending in zero is divisible by 2. One way to show this is by example, or a number of examples. But should we leave it an inductive logic?
We could argue from long division, assuming students are reasonably fluent in both short and long division. When you are dividing by 2 you will have a remainder of either zero or 1 at each step in the process. If the remainder is 0 at the next to the last step, we are done. Bring down the 0 and the remainder is 0. If the remainder is one at the next to last step, bringing down the final zero gives 10, which is divisible by two, so there is no remainder. Again we are done.
For this explanation we are dividing the proof into case 1 and case 2. Dividing the proof into a number of cases happens many times in math. Is it worth while to discuss this with elementary students? I don't know.
Can elementary students understand that "any number ending in 0 is divisible" is a theorem, not an axiom? I'm not sure. To distinguish between axioms and theorems may be asking too much for this age. Can they understand the difference between inductive and deductive logic? My experience at trying to teach this distinction to college freshmen makes me very pessimistic on that point.
Now I want to skip to divisibility by 9. Once divisibility by 9 is understood, divisibility by three can be related as an easy extension.
The rule for divisibility by 9 is simple. Add up the digits of the number. If the sum of the digits is divisible by 9, then the entire number is divisible by 9. For example we can know that 783 is divisible by 9, because the sum of the digits, 7 + 8 + 3 = 18, which is divisible by 9. Students can learn the rule and apply the rule fairly easily. But can the understand why the rule works? Do most teachers understand why the rule works? I can't really answer that question, but I would take a guess that probably many teachers do not. But it is not too hard to understand, especially having established the principles that we have. First we need a bit of preliminary work.
Ten of anything can be seen as nine of those things plus one of those things. As an example, ten of the number 38 can be seen as nine 38's and one more 38. This is the distributive law of course. If the students have already been exposed to the distributive law it should make sense. If they have not been exposed to the term "distributive law" it should still make sense intuitively. Indeed multiplication has been developed partly on the basis of the distributive law, intuitively if not explicitly. How do we know that six eights are forty eight? We can draw objects, perhaps x's. Eight groups of six x's each will count up to 48 every time. However we can also arrive at the figure of 48 another way. If we know that five eights are forty, then we can simply add one more eight to five eights, add another 8 to 40 and we have 48 for six eights. Five eights and one more eight make six eights, just as five apples and one more apple make six apples, or five dozen and one more dozen make six dozen.. Hopefully that sort of thing has been done many times in learning multiplication.
To continue with the example of 380:
So far in this example we have done nothing about divisibility. We will get there in a moment. At this point just notice that the first number in step 2, 9 ´ 38, is divisible by 9 because it "contains" 9, or because rule number one applies.
There is a bit more preliminary work to do. One hundred of any number can be seen as 99 of that number plus 1 more of that number.
In step 4 we have two numbers. The first number is 99 ´ 4. Is that divisible by 9? It is, because 99 is divisible by 9. It is 9 times 11. So by rule 1, (99 ´ 4) is divisible by 9. We can say that it "contains" 9.
But still we have not arrived at a conclusion of divisibility by 9 for a number that we start out with. To do this let us take a slightly different number, 486. Is 486 divisible by 9? The rule says to add 4 + 8 + 6. That is 18. 18 is divisible by 9, therefore by the rule the whole number, 486, is divisible by 9. It can be checked out by dividing. The rule works, but can students understand why? Here's the explanation:
Finally in step 6 we have two numbers. Therefore we can apply rules 1 and 2. Can students understand that (99 x 4 + 9 x 8) is one number? I presume they can, perhaps with some careful explanation. However I don't have any experience to base this on. If they can, and then they can see that (99 ´ 3 + 9 ´ 8) can be thought of as two numbers, and each of those numbers are divisible by 9, because they each contain 9, and by rule 2 the sum of those two numbers must therefore be divisible by 9, and then rule two says that the two numbers in step six will be divisible by 9 providing the second number, (4 + 8 + 6) is also divisible by 9.
Then all that is left is to show that the second number (4 + 8 + 6) is simply the sum of the digits. Put it all together and you have the rule for divisibility by 9.
Here's another example, this time with a 5 digit number. Is 46738 divisible by 9?
So again in step 3 we have two numbers. The first number is divisible by 9 because every part of it contains 9. The second number again is simply the sum of the digits. If it is divisible by 9, then rule 2 applies and 46738 is divisible by 9.
Is step 1 a problem for students? Do they understand where the 10,000, the 1000, and so on come from? I would think this is just a matter of understanding place value and that this understanding has been developed previously. It would seem to be needed for carrying and borrowing, for multiplying and dividing, and for understanding decimals. Perhaps it would need some additional explanation at this point. If they do not understand place value then of course nothing else in this explanation would make any sense.
The divisibility rule for 3 is the same as for nine. Add the digits. If the sum of the digits is divisible by 3, then the number itself is divisible by 3. Why this is so can easily be seen. Is 38735 divisible by 3?
Again in step 3 we have two numbers. In the first number (in the first set of parentheses), every addend has 3 has a factor, so the number is divisible by 3. The second number is once again simply the sum of the digits. If the sum of the digits is divisible by 3, then again rule 2 applies. In this case the sum of 3 + 8 + 7+ 3 + 5 is 26, not divisible by 3, so 38735 is not divisible by 3. We can simplify the rule just a bit by "casting out 3's". Instead of adding 3 + 8 + 7+ 3 + 5, cast out the 3's and just add 8 + 7 + 5, to get 20, which is not divisible by 3 and tells us that 38735 is not divisible by 3.
I think the above is understandable to students who know arithmetic pretty well. Algebra notation makes things even more plain, but only if one is reasonably fluent in using algebra notation, which elementary school students are not expected to be. But I will go ahead and use some algebra notation for whatever it is worth.
Unfortunately we must begin by breaking a rule of algebra notation. In normal algebraic notation xy means the product of x and y, which we could write as x ´ y. Using the cross for multiplication makes a lot of sense in arithmetic. It is a symbol in very widespread use. However it presents a problem in algebra. We traditionally use the letter x a lot in algebra to represent an unknown quantity, and the x is very easily mistaken for the cross of multiplication, and vice versa. Therefore we normally don't use the cross in algebra to mean multiplication. Instead we just omit any symbol for multiplication. Thus xy means the product of x and y, and ab means the product of a and b, and 3w means the product of three times w. If we try to write the number 4 ´ 5 with this rule we would get 45, which would be interpreted as 45. We can write (4)(5) to mean 20. Or we can write 4(5), or (4)5.
In standard algebraic notation "cba" would obviously mean c times b times a. But for the moment I am going to break that rule. We need a way to represent a three digit number, or a five digit number, or a two digit number. Therefore, in this context, when I speak of the number cba, I will mean c hundreds plus b tens and a ones. Or if I speak of the four digit number dcba, I will mean d thousands plus c hundreds plus b tens plus a ones.
Example: The number 49,382 can be represented as edcba, where e = 4, d = 9, c = 3, b = 8 and a = 2, because 49,382 = 10000(4) + 1000(9) + 100(3) + 10(8) + 2.
Now the divisibility rule for 9 can be expressed as follows:
The number dcba is divisible by 9 if and only if d + c + b + a is divisible by 9 .
This is because the number dcba means 1000d + 100c + 10 b + a, and
Apply the basic principle, rule number 1, and rule number 2, and we have the rule for divisibility by 9. The rule for divisibility by 3 can be expressed similarly. Perhaps there is a better way to express this. This formulation has the fault of using conflicting notations, but it has the virtue of being algebraic and concise.
With this notational convention in place we will now return to the remaining rules for divisibility. The divisibility rule for four is pretty simple. You just look at the last two digits. If the last two digits are divisible by four then the whole number is divisible by four. Thus the number 43,652 is divisible by four because 52 is divisible by four.
I expect most kids will understand this intuitively, providing they have been adequately taught in arithmetic. But having established the three principles about divisibility we ought to apply them the rule for divisibility by four. That is very simple. Using our example above, we can decompose 43,652 into the two addends 43600 and 52. By rule two the whole number is divisible by four if both addends are divisible by four. 52 can be shown to be divisible by four by simply doing the division. 43600 can be shown to be divisible by four simply by dividing. But can we do more than that? Perceptive students will intuitively figure out that any hundred is divisible by four. Can we make that into a theorem? Can students studying arithmetic approach it as a theorem? All that would mean is realizing that any hundred is divisible by four, and figuring out some way to make that appear sensible and obvious. I would suggest something like the following:
In line 6 of this figure there will be a remainder of 0, or 1, or 2, or 3. Thus there are four cases in the "proof". It is easy to show that in each case the division comes out even. Therefore any number of hundreds is divisible by 4 without a remainder. Thus we have a theorem. We can state that theorem as "Any number of hundreds is divisible by 4." Or we can state it as "If a number ends in 00 then it is divisible by four." I would think that at this state, elementary school, exact wording would not be very important. The idea is important. And I would think that at this stage students could understand that this rule, however we word it, is a logical consequence of what we have said and done before. Therefore I would think that the use of the term "theorem" would be appropriate and meaningful to students. However I have no experience on which to base that.
Are students at the elementary school age quick to recognize the multiples of four up to 100? Are they quick to recognize leap years or election years? I don't have the experience to know this. Would it be sensible to bring up leap years and election years in class? I would think it would be, but again that would depend on whether or not there is enough time available.
Divisibility by eight can be handled similarly, but it is the thousands, not hundreds that are divisible by 8. How are the students to recognize and remember all the multiplies of 8 up to a thousand? Perhaps it is worth while to spend time on this, or perhaps not. If not, then the method involves dividing the last three digits by eight.
Divisibility by six can easily be done by a double check. First check for divisibility by two, and then by three.
Divisibility by ten is easy, any number than ends in zero is divisible by ten. I suppose there are various ways in which that can be explained and discussed if it is not already obvious to students. Again I would think that the use of the term "theorem" would be appropriate and understandable.
We have now covered all single digit divisibility rules except for 7. Seven is a little harder. First we will start with the rule. The rule is not as simple as we would like, but it's not really hard either. And it is recursive. For large numbers it may need to be applied a number of times. Take the one's place digit , double it, and subtract it from the number that results by truncating the one's digit from the rest of the number. (The ten's digit becomes the one's digit.)
Example: Is the number 4562 divisible by 7? Double 2 to get 4, then subtract 4 from 456 to get 452. If 452 is divisible by 7 then 4562 is divisible by 7. But it is not obvious whether or not 452 is divisible by 7, so apply the rule again. (That's what we mean by "recursive") Double 2 to get 4, then subtract 4 from 45 to get 41. 41 is not a multiple of 7, therefore the original number, 4562 is not divisible by 7.
Maybe 4563 is divisible by 7. 4563 ® 456 - 6 = 450 ® 45-0 = 45, not divisible by 7.
Maybe 4564 is divisible by 7. 4564 ® 456 - 8 = 448 ® 44 - 16 = 28, so 4564 is divisible by 7.
Here I am using the arrow as a general symbol to indicate the next step. This might be a good place to point out that it would definitely be wrong to put an equals sign in place of that arrow. It is definitely not true that 4563 = 456 - 6. And equals sign should not be used indiscriminately to mean the next step, as algebra students will sometimes do.
I presume kids, or adults, can learn to apply this rule in a few minutes. But why in the world does it work? It certainly seems strange. Can anyone reasonably expect to understand why it works?
Actually it is not too complicated, and I think kids can understand it when it is explained well. I do not expect kids will figure it out on their own. I struggled with it for the better part of two days and got nowhere. Then I find the explanation online. First we need our usual form of notation. A number such as 4564 can be expressed as dcba, where d = 4, c = 5, b = 6, and a also = 4. Then the number can be written algebraically as
1000(4) + 100(5) + 10(6) + 4, because in the present context dcba means 1000d + 100c + 10b + a. However at the moment we need a slightly different way to write this.
4564 = 10(456) + 4. I will write this as 10b + a.
It is the 10b + a that we are interested in. It turns out that if 10b - a is divisible by 7, then b - 2a is divisible by 7. What in the world should that be? And how can we show that indeed it is. We can make it seem a little more concrete with a table such as the following:
Note that the remainder when dividing by 7 of 10b + a and of b - 2a will not be the same, unless the remainder is zero. But every time 10b + a is divisible by 7, then b - 2a will be divisible by 7. Therefore the rule will always work. But why should it? Is there some algebraic manipulation that will get us from 10b + a to b - 2a?
Algebraic manipulation is not needed. What is needed is the three principles that we have already established, the basic principle of what we mean by divisibility, and rule 1, a number is divisible providing one factor of that number is divisible, and rule two, a number is divisible providing both addends are divisible. These principles tell us that if a number, n, is divisible by 7, then the number n -21x is also divisible by 7. x can be any integer and this will be true. But not just any integer x will be relevant. What is relevant is that we choose for x the a from the expression 10b + a. Then a little algebraic manipulation gives us:
If the number n is divisible by 7, then n -21a is divisible by 7, because of rule 2, and 10(b-2a) must be divisible by 7, because 10(b-2a) is the same number as n - 21a. Obviously 10 is not divisible by 7, so by rule one b-2a must be divisible by 7. That this is true is illustrated by the chart we made above.
Can elementary school students understand all this? I don't know. If they can, how much time would it take, and is that time available? I don't know.
So what is the place of divisibility in the teaching of arithmetic? I must leave that to others to figure out. If there is time to do everything Iíve discussed in this article, and if students can understand it all, at least on an intuitive or concrete level, it would seem that there is much to gain.