The Power of Algebra

In this session we look at a novel way to do division by dividing by each of two smaller factors and then subtracting the results.

Teachers’ Notes

Background: The problem here was actually first discovered by a Year 6 student. She noticed that if you want to divide something by 12 you can first divide it by 3 and then by 4. The answer is the difference between these two divisions.

Over the course of the next two sessions we will investigate this with a view to first proving that it is true and then second showing how to generalise it. We will prove these results by using algebra. These proofs show the power of algebra. We first take the argument that shows why the girl’s observation is correct in some special cases. Then we replace these special cases by 12n, effectively replacing the special multiples of 12 by an unknown n. The neat thing is that exactly the same steps show that the result is true for any multiple of 12.

We follow this up by seeing when the girl’s observation works. What was so special about 12? From here we can go to the generalisation that we give below that tells what happens for any number that is a multiple of any two numbers a and b. The proof for the special cases works in the general case again. Once again we can see how powerful algebra is. In one go it can do an infinite number of cases and so prove that a result holds for all multiples of any numbers a and b.

Proof 1: Let N = 12n. The girl’s observation says that N/12 = N/3 – N/4. Well first of all, N/12 = n. Then N/3 = 12n/3 = 4n and N/4 = 12n/4 = 3n. So N/3 – N/4 = 4n – 3n = n = N/12 and the result is proved.

The point to be made here is that if we took n = 13, say, the process above would show that 156/12 = 156/3 – 156/4. The algebra enables us to do every case in a few lines. We don’t have to do an infinite number of special cases.

Generalisation: To see what is actually going on here let’s look at a special case. Suppose that we look at multiples of 21. How can we find what happens when we divide by 21? Since 21 = 3 x 7 perhaps we could divide the number first by 3 and then by 7 and subtract the two answers. In fact it doesn’t quite work that way. Look at an example. Divide 252 by 21 and we get 12. Divide 252 by 3 and we get 84 and by 7 and we get 36. But 84 – 36 = 48 ≠ 12. However, 48/4 = 12 and 7 – 3 = 4. And that’s what’s going on. 252/21 = (252/3 – 252/7)/(7 – 3). This was actually the same thing that was going on in the girl’s observation but it wasn’t so obvious because 4 – 3 = 1. What she was actually doing was N/12 = (N/3 – N/4)/(4 – 3).

Suppose that a and b are two numbers and we are interested in using the girl’s method to calculate N/ab, where N = abn, for any number n. Then (N/a – N/b)/(b – a) = (bn – an)/(b – a) = (b – a)n/(b – a) = n = N/ab.

This general proof is exactly what you would do with a special case.  Check it out with N = 252 and a = 3 and b = 1.

Teaching Sequence

 

1. The following observation was made by a girl in Year 6.  If you want to divide a number by 12, you can divide it by 3, then divide the first number by 4, and then subtract the second answer from the first.
   In a whole class situation, get the class to consider this observation.
   Is the observation correct?
   How can we find out?
   How can we justify it as a rule?

2. At this point you might like to let the class work on some examples of numbers that are divisible by 12 to see if the girl’s observation is correct. Let them use their calculators to produce at least three numbers of their own that have 12 as a factor.

3. Let different members of the class report on their findings. All of them should support the girl’s observation.
   It looks as if we have a good conjecture here.
   How can we prove the girl’s conjecture?

4. Discuss how they could justify the conjecture. Get them to see how to explain it in special cases. For example, 12 into 156 goes 13 times; 3 into 156 goes 52 times; 4 into 156 goes 39 times; 52 – 39 = 13. But we could look at this in another way. 156 = 12 x 13 = 3 x 4 x 13. So 156/3 = 4 x 13 and 156/4 = 3 x 13. What’s more 156/12 = 13 = 4 x 13 – 3 x 13 = 156/3 – 156/4.

5. Repeat this for other numbers that are divisible by 12. For instance, 252. Doing exactly the same as we did above, 252 = 12 x 21 = 3 x 4 x 21. So 252/3 = 4 x 21 and 252/4 = 3 x 21. Then we see that 252/12 = 21 = 4 x 21 – 3 x 21 = 252/3 – 252/4.

6. Get the class to repeat this sequence of calculations that show that the conjecture is true for several numbers. Then tell them that if we were going to prove the conjecture for all numbers divisible by 12, we would have to do that calculation for every multiple of 12. That would certainly take a long time!
   How can we do it in a short time?

7. The answer is by algebra. All we have to do is to replace the 13 (or the 21) or whatever multiple of 12 they have been using) by the letter n. So every number that is divisible by 12 can be written as 12n. But 12n = 3 x 4 x n. So 12n/3 = 4n and 12n/4 = 3n. Then we can see that 12n/12 = n = 4n – 3n = 12n/3 – 12n/4. This is exactly the same justification that we used for the two particular cases above. But the justification using n works for every number n and so proves the conjecture for every multiple of 12. The girl’s conjecture is true.
   Give them the chance to copy this proof into their books.

8. Note that, using algebra, we have turned an argument that worked for special multiples of 12 into an argument that works for all multiples of 12.

9. Does this only work for 12?
   If you want to divide a number by 20, can you divide it by 4, then divide it by 5, and then subtract the second answer from the first?

10. Let them work on this problem in groups. The faster groups can be challenged to think about any divisor that is the product of two numbers. Can they find any counterexamples to the conjecture that if you want to divide a number by a x b, you can divide it by a, then divide it by b, and then subtract the second answer from the first?

11. Give the groups a chance to report back on what they have found. Discuss what they have produced. Are there any bigger conjectures here?

Session 2

In this session we justify the generalisation that we get from the method of the last session. We also see how to extend the method to factors that differ by more than 1.

Teaching Sequence

1. Remind the students where they got to in the last lesson.  There they found that one way to divide by 12 was to divide first by 3 and then by 4 and then find the difference between the two answers.  But the same thing happens for division by 20 if you divide first by 5 and then by 4 and then subtract.

2. In a whole class situation ask one of the following questions, give them time to answer, and then discuss what they have found. Then go on to the next question.
   Does this method work for 20 if we divide first by 2 and then by 10?
   Does the method work for 30 if we divide first by 5 and then by 6?
   What about 30 if we divide by 3 and then by 10?
   When do you think the method works?
   Can you make a conjecture? (Works if the factors differ by 1.)

3. Let them go into their groups to look at more situations.
   Is there a more general result?
   For the faster students you might like to ask them if they can justify the conjecture using a method similar to the one used with 12 that they copied into their book in the last lesson.

4. Bring the class back together and discuss the results that they have found. Get them to show how the proof from the last lesson for 12 can be changed to give a proof for 20, then 30, then any specific number. Ask them if they can extend this to the number a(a – 1).

5. Let them copy the particular proofs for 20, 30 and some other number. Ask them if they can predict how the method will go for (i) 35 – dividing first by 5 and then by 7; (ii) 80 – dividing first by 5 and then by 16; and (iii) 40 – dividing first by 5 and then by 8.

6. Go around the groups to see how they are progressing. As they get through those three examples ask them if they can conjecture a generalisation of the method. Can they prove this conjecture?

7. Discuss what the groups have found. Consider any proofs that they have. Have them write up their data and their proofs in their books.

Session 3

In this session we justify conjectures in two problems arising from taking a 2-digit number and reversing its digits.

Teachers’ Notes

The numbers that we give in brackets next to the problems here are the number that they are given in Discovery, Level 6.

New Problem (1): Take any 2-digit number; reverse the digits; subtract the smaller number from the larger number. What can you say about the result?

Let’s do a few examples.

61 – 16 = 45;                  81 – 18 = 9;                 92 – 29 = 63;                  53 – 35 = 18.

Each of these numbers is a multiple of 9. Is that always the case?

Let’s first check out why particular numbers give you a multiple of 9 and then show that any 2-digit number will give you a multiple of 9.

61 – 16 = 6 x 10 + 1 – 1 x 10 – 6 = 6(10 – 1) + 1(1 – 10) = (10 – 1)(6 – 1) = 9 x 5.
87 – 78 = 8 x 10 + 7 – 7 x 10 – 8 = 8(10 – 1) + 7(1 – 10) = (10 – 1)(8 – 7) = 9 x 1.
92 – 29 = 9 x 10 + 2 – 2 x 10 – 9 = 9(10 – 1) + 2(1 – 10) = (10 – 1)(9 – 2) = 9 x 7.
52 – 25 = 5 x 10 + 2 – 2 x 10 – 5 = 5(10 – 1) + 2(1 – 10) = (10 – 1)(5 – 2) = 9 x 3.

What we have just done can been turned into a proof if we use a bit of algebra. Suppose that we have the 2-digit number ab, where a is bigger than b. We’ll use exactly the method we used with the four cases above to prove that ab – ba is divisible by 9 no matter what the values of a and b.

ab – ba = a x 10 + b – b x 10 – a = a(10 – 1) + b(1 – 10) = (a – b)(10 – 1) = (a – b) x 9.

Comment 1: Here we see the power of algebra again. Using the same method that we used for special numbers we are able to prove something is true for every 2-digit number.

Comment 2: One of the things that mathematicians like to do is to see if they can push a method to do more than they had expected to do. In this example we set out to prove that ab – ba was divisible by 9. Not only have we succeeded but we have done more. In the proof above the final answer was (a – b) x 9. So not only do we know that the answer is a multiple of 9 but we also know exactly what multiple it is: a – b!

Comment 3: Some students might think that if you take 44 – 44 and get 0, that this is not a multiple of 9. But 0 = 9 x 0. So the proof that we have found works equally for two different digits in the 2-digit number and when the two digits are the same.

Another problem (2): OK so we know that 61 – 16 = 5 x 9 but what happens if we now reverse the digits of this answer and add the new number and the old answer together. Let’s do it for the four examples we have above.

61 – 16 = 45; reverse 45 to give 54; 45 + 54 = 99
87 – 78 = 9; reverse 9 to give 90 (a bit of poetic licence here); 9 + 90 = 99.
92 – 29 = 63; reverse 63 to give 36; 63 + 36 = 99.
52 – 25 = 27; reverse 27 to give 72; 27 + 72 = 99.

but 44 – 44 = 0; reverse 0 to get 0; 0 + 0 = 0.

So it looks as if there is going to be more than one answer to this reversing and adding problem. It’s certainly not always going to be 99. But maybe we can conjecture for the moment that there are going to be only two answers, 0 and 99.

Conjecture: Let ab be a 2-digit number and let ab – ba = M. Reverse the digits of M to give N. Then M + N is either zero or 99.

Another proof: Let’s prove the conjecture.

If a = b, then ab – ba = 0. Reversing zero gives zero. Adding zero to zero gives zero. Having got rid of that, we can get down to the serious cases. From now on we can assume that a is bigger than b.

Consider the 2-digit number ab, where a is greater than or equal to b. Now to be able to reverse the answer to ab – ba we have to be able to find the digits in the subtraction. What is the units digit? b – a? No, because this number is negative. So we would have to ‘carry over’ a 10 to help things out. The units digit is therefore 10 + b – a. (To check this out see what happens in the case where a = 6 and b = 1.)

So ab – ba = some multiple of 10 plus (10 + b – a).

What is the multiple of 10? It is nearly a – b. But remembered that we ‘carried over’ a 10 to help out the b. So the tens digit is a – b – 1. This gives us

ab – ba = (a – b – 1) x 10 + (10 + b – a).

Reversing this last number gives (10 + b – a) x 10 + (a – b – 1).

Adding the two numbers gives [(a – b – 1) + (10 + b – a)] x 10 + [(10 + b – a) + (a – b – 1)] = [10 – 1] x 10 + [10 – 1] = 99!

So the conjecture was true.

Teaching Sequence

1. With the whole class together, tell the students the new problem about reversing the digits of a 2-digit number and subtracting the two numbers.
   What happens if we take 61, reverse it and subtract the smaller number from the larger number?
   Repeat for other 2-digit numbers.
   List the answers on the board.
   What do you notice about these numbers?
   Do we have a conjecture?
   Have a discussion about whether there is a counter-example or a proof. Don’t pressure them for the particular multiple of 9 at this point.

2. Let them work in their groups on the conjecture by trying more examples. Try to move the more able students to think abut a proof. Lead them to look at the proof of some special cases as we have done in the Teachers’ Notes.

3. Bring the groups together. Discuss what they have done.
   Try to get a proof of the conjecture. If they can find a proof that is different from the one here or the Proof by Exhaustion in Discovery, Level 6, so much the better.  Otherwise try to get to a proof via the proof of several special cases. Notice how algebra will take care of every case in one go.
   Be sure to note what the multiple of 9 is in each case and how this comes out of the general proof.
   Give the students time to write this up in their books. Include the examples they found, the proof in specific cases, and the proof in general.

4. Repeat the process with the reversing and adding problem.
   The conjecture here is just as easy to see but is much more difficult to prove. The ‘carry over’ part is especially difficult. Do this proof with several special cases before you try the general case.

5. Give the students time to write up the results in their books.

Session 4

In this and the next session we extend the problems of the last session to 3 digits

Teachers’ Notes

Yet another problem: What is the natural extension of the 2-digit problem that we considered in the last session? Perhaps we should ask: what are the natural extensions of the 2-digit problems that we considered in the last session?

Problem 1 (3). Take a 3-digit number. Reverse the digits. Subtract the smaller number from the larger one. What answers do you get?

Solution: As usual with a problem like this the first thing to do is to experiment. (There’s certainly no formula that will give the answer to you.) So let’s try some numbers at random and see what we get.

563 – 365 = 198;         815 – 518 = 297;         783 – 387 = 396;      571 – 157 = 414.

That hasn’t quite exposed all of the possibilities. May it’s possible to get a 2-digit answer. How could we do that? 564 – 465 = 99. That’s one possibility. And could we get a 1-digit answer? Hmm. We can do that if the answer is zero but it’s hard to see how else we could manage it. Zeros come if we use palindromes.

Let’s stick with the 3-digit answers for a minute. What do the four answers above have in common? If you recall Guzzinta, Level 6, you’ll realise that they are all divisible by 9 and by 11. Can we conjecture that all of these differences are divisible by 99? Well, the one 2-digit answer that we got was divisible by 99 and so is zero. So that seems as reasonable a conjecture as any at the moment.

Can we develop the special cases above so that we show the factor of 99 and at the same time set up a nice algebraic proof of our conjecture? Let’s try.

563 – 365 = 5 x 100 + 6 x 10 + 3 – 3 x 100 – 6 x 10 – 5.

Well one nice thing that happens here is that we lose the 60 and we only have to worry about the 5s and the 3s. So

5 x 100 + 6 x 10 + 3 – 3 x 100 – 6 x 10 – 5 = 5(100 – 1) + 3(1 – 100) = 99(5 – 3) = 99 x 2.

That ought to just about do it. All of the other special cases will go exactly the same way. It’s just like the ‘reverse’ problem in Sessions 2 and 3.

And again we get a bonus. We now not only that the answer is a multiple of 99 but we also know that the multiple is the difference between the hundreds and the tens number!

In general though,

abc – cba = a x 100 + b x 10 + c – c x 100 – b x 10 – a = a(100 – 1) + c(1 – 100) = 99(a – c).

Problem 2 (4). Take a 3-digit number and reverse the digits. Subtract the smaller number from the larger number. Call the result N. Reverse the digits of N to form M. Add M and N. What answers do you get?

Solution: Again we start with experiments.

563 – 365 = 198 and 198 + 891 = 1089;      815 – 518 = 297 and 297 + 792 = 1089

That looks ominous! Do you always get 1089?

Can we set up an algebraic proof?

783 – 387 = 396 and 396 + 693 = 1089 but that hasn’t set anything general up. How can we do that? Let’s try again and have in mind what we did in Session 3.

783 – 387 = 7 x 100 + 8 x 10 + 3 – 3 x 100 – 8 x 10 – 7 = 7 x 100 – 3 x 100 + 3 – 7.

And it’s here that we can’t take 7 from 3 so we carry over a 10. So we might want to write 3 – 7 as 10 + 3 – 7 but we’ll have to remember that we did the ‘carry over’.

So 783 – 387 = 7 x 100 – 3 x 100 – 10 + [10 + 3 – 7].

But now we are going to need a ‘carry over to allow us to subtract the 10.

So 783 – 387 = 7 x 100 – 3 x 100 – 100 + 100 – 10 + [10 + 3 – 7] = 7 x 100 – 4 x 100 + [90] + [10 + 3 – 7] = [7 – 3 – 1] x100 + 9 x 10 + [10 + 3 – 7].

Now reverse the digits and add.

{[7 – 3 – 1] x 100 + 9 x 10 + [10 + 3 – 7]} + {[10 + 3 – 7] x 100 + 9 x 10 + [7 – 3 – 1]} = {[7 – 3 – 1] + [10 + 3 – 7]} x 100 + {9 + 9} x 10 + {[10 + 3 – 7] + [7 – 3 – 1]}.

Fortunately that all simplifies to give

[10 – 1] x 100 + 19 x 10 + [10 – 1] = 9 x 100 + 1 x 100 + 9 x 10 + 9 = 1089!

And that can all be algebraified just by replacing 7 by a, 8 by b and 3 by c. (See Session 2 of Proof, Level 6 for a complete justification.)

Well not quite. There is a problem that we need to concern ourselves with. What if a = c? In what we did above we implicitly assumed that a was bigger than c. That was why we did the ‘carry over’. If a = c on the other hand, we get zero as the difference between the two numbers abc and cba so when we reverse and add we still get zero. But that covers all possibilities. So the answer here is 1089 unless we started off with a palindromic number.

Teaching Sequence

1. Remind the students of the two problems we did in the last two sessions. Discuss how you went about collecting evidence for the conjectures and how you finally solved the problems. You might want to recall how they worked on special cases and how these special cases were turned into an algebraic proof.

2. Talk about the way mathematicians like to extend problems. There are lots of examples of this in the Problem Solving section of the site and in the Bright Sparks problems.
   How can we extend the problems that we have been working on?
   What is the next step that we might try?
   Hopefully they will be able to think of extending the problem to three digits. Get them to word the problems as carefully as they can.

3. Start on Problem 1 (difference only) by getting them to try a selection of random 3-digit numbers to see what sort of differences occur. It might be useful to put several examples on the board after they have found them. Perhaps even let the students put the numbers on the board.

4. Next consider the data that you have and see if they can come up with some conjectures. Discuss how you might prove a conjecture in the special case of a particular 3-digit number.

5. Let the students go into groups to work on other special cases. Hopefully they could then use their previous notes to help them produce a general proof. Go around the groups to provide help and ask leading questions.

6. Bring the class together and let them tell you what they have discovered. Let them argue their cases on the board. Then give them time to write up the proof in their books.

7. Now consider Problem 2. Go through the same teaching sequence as above. This may spill over into the next session

Session 5

Here we look at an extension of the problems of the last sessions.

Teachers’ Notes

Problem 3 (9). Some students might suggest that instead of going to three digits you use sums instead. In other words take a 2-digit number, reverse the digits and add. Is there any pattern here?

Solution: Here with the 2-digit number ab, you get

ab + ba = 10a + b + 10b + a = 10(a + b) + (a + b) = 11(a + b).

Problem 4 (10). Now after reversing, reverse again and add the two numbers you get. In other words, start with ab and then add to ba to give a number A. Then reverse the digits of A to give B. Add A and B to give C. What can you say about

Solution: The difficulty here is keeping track of the digits. Now A = ab + ba = (a + b) x 10 + (a + b) but a + b may be bigger than 10 and we may get some carrying over. To keep track of this we do things slowly. In other words we’ll break the rest of this argument down into cases. First we’ll let a + b = c, where 0 ≤ c ≤ 9, and then we’ll let it equal d + 10, where 0 ≤ d ≤ 8. Note that this covers all cases because the sum of any two digits is strictly between 0 and 18.

Case 1: A = cc = B, so C = 2cc = 2c(10 + 1) = 22c. Because c has ten values, we’ll get 10 values for C this way.

Case 2: A = (d + 10) x 10 + (d + 10) = 100 + (d + 1) x 10 + d. Because d has nine values, we’ll get 9 values for C this way.

Just to see what we get we’ll list them all.

Problem 5 (10). Repeat Problems 3 and 4 with 3-digit numbers.

It turns out that abc + cba is not in general divisible by 11. However, abcd + dcba is. There seems to be a parity thing operating here. If the number of digits in the original number is even, then reverse and add gives a factor of 11. If the number of digits in the original number is odd, then reverse and add may not give a factor of 11.

Can you find all of the 3-digit examples where 11 is a factor? It turns out that you need 2(a + c) + 9b to be divisible by 11. This comes directly from abc + cba = (a + c) x 101 + 20b after removing 99(a + c) and 11b.

Pushing this a little further we can show that this factor of 11 only appears if and only if a + c – b is divisible by 11. (In other words, from Guzinta, Level 6, if and only if abc is divisible by 11.) To see this first note that

P = abc + cba = (a + c) x 101 + 20b = 11[9(a + c) + b] + 2(a + c) + 9b = 11[9(a + c) + 2b] + 2(a + c) – 2b.

Now P is divisible by 11 if and only if 2(a + c) – 2b is. But since 2 is not a factor of 11, 2(a + c) – 2b is divisible by 11 if and only if (a + c) – b is. But abc is divisible by 11 if and only if (a + c) – b is.

Finally reversing the digits of P and adding the new number to P is also problematic. There clearly must be a limited number of answers (there are only a thousand starting numbers) but how to define these neatly (as, for example, in the Solution to Problem 4 above), is far from obvious. Let us know when your class solves this one.

Teaching Sequence

1. Repeat the approach of the sessions above with the rest of problems from Session 4.

2. What other problems can you come up with by extending the problems of this unit? What answers can you get?