Int this unit look for patterns in number problems and see how far we can extend basic ideas. The main point of this unit is to produce conjectures. Some time is also spent on developing proofs.
This unit illustrates the way that pure mathematics works; from posing a problem, to experimenting, to conjecturing, to proving, and to extending and generalising. Therefore, it employs general strategies that can be used time and time again. Although we use these strategies in finding patterns in number problems they can be used to introduce new ideas in any area of Mathematics. In fact you will find these strategies at least implicitly throughout the majority of the units in this web site.
The key element of this unit is looking at special cases in order to come up with a guess at a pattern. These guesses are called conjectures.
We do mention here the method of Proof by Exhaustion. This method allows us to prove something by looking at every case. This method is covered in detail in the Teachers’ Notes in the early sessions of the unit.
The learning opportunities in this unit can be differentiated by providing or removing support to students and by varying the task requirements. Ways to differentiate include:
This unit is focussed on investigating Conjectures and Proof by Exhaustion, and as such is not set in a real world context. You can increase the relevance of the learning in this unit by providing ample opportunities for students to create their own problems, create their own representations of a task, and participate in productive learning conversations.
Te reo Māori kupu such as tauira (pattern), tāiringa kōrero (conjecture), and hāpono (proof) could be introduced in this unit and used throughout other mathematical learning.
In this session we look at the basic two problems of this unit and see what progress can be made with them.
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?
At this point we suggest that you draw up a table. You may not bother with the last column at first. There is no reason to ask that question until you get further into the problem.
ab | a | b | Multiple of 9 | What multiple? |
61 | 6 | 1 | yes | 5 |
81 | 8 | 1 | yes | 7 |
92 | 9 | 2 | yes | 7 |
53 | 5 | 3 | yes | 2 |
Comment: Some students might think that if you take 44 – 44 and get 0, then this is not a multiple of 9. But 0 = 9 x 0. So it looks as if every 2-digit number gives us a multiple of 9.
Conjecture 1: If we take any 2-digit number; reverse the digits; subtract the smaller number from the larger number, then the answer is a multiple of 9.
But we can go further.
Conjecture 2: The multiple of 9 in Conjecture 1 is the difference between the two digits of the original 2-digit number.
It’s worth noting that Conjecture 2 includes multiples of zero.
Problem 2: Take any 2-digit number; reverse the digits; subtract the smaller number from the larger number to give the number M. Now reverse the digits of M to give the number N. Add M and N. What answer do you get?
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 (use 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.
Oh but 44 – 44 = 0; reverse 0 to get 0; 0 + 0 = 0.
This might be a time to draw up another table.
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 only going to be two answers, 0 and 99.
Conjecture 3: In Problem 2, the answer will either be 99 or 0.
So far, the conjectures we have can all be proved by Exhaustion (i.e. by exhausting all cases). We can prove Conjectures 1 and 2 by looking at all of the 90 2-digit numbers to see what answer they give.
In actuality, when you think about it, it’s not that many. After all we only have to worry about the 2-digit numbers ab where a ≥ b. Now that’s how many? 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 54. This summation can be done quickly using Gauss’ method – as described in the Algebra Information on this site. This comes from 2 for a = 1 (10 and 11), 3 for a = 2 (22, 21 and 20), etc. If your class takes 2 of these each you should be able to knock over a proof by covering all cases in about 10 minutes!
Conjecture 3 only requires us to look at all multiples of 9 up to 81. There are only 10 of these (including 0; the rest are 9, 18, 27, 36, 45, 54, 63, 72, and 81). That shouldn’t take as long as the other two Conjectures. We could start with 00 + 00 = 0; 09 + 90 = 99; 18 + 81 = 99; and so on.
Let them work in their groups on filling in more entries on the table. Get them to think about what conjectures they could make.
What patterns (conjectures) can you find? (See Conjectures 1 and 2.)
Let different groups report on their findings to the whole class.
Discuss what they report.
Let them write their conjectures in their books.
Be sure to make students aware that we don’t know if these conjectures are true or not. All that we can say at this point is that we have found no counter-examples so they are only probably true. Encourage the more knowledgeable students to think about how they might prove the conjectures that have been made. Proof by Exhaustion is one approach (see Teachers’ Notes).
Is that feasible here?
How many calculations will be needed?
How many 2-digit numbers are there? (They may think that there are 100 but let them discover that there are in fact only the numbers from 10 to 99 so this is 100 – 1 – 9 = 90.)
Do we have to test all of these numbers? (In fact we only have to worry about 2-digit numbers ab, where a ≥ b. How many of these are there?) You might want to revise the sum of the first n whole numbers here.
Is it feasible for us to divide these up between us?
In this session we extend Problems 1 and 2 and try to make conjectures about their answers.
One of the things that mathematicians try to do is to extend and generalise problems. In this session we extend the two problems of Session 1. You can lead them in this by posing Problems 3 and 4.
Problem 3: Take any 3-digit number; reverse the digits; subtract the smaller number from the larger number. What can you say about the result?
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; 751 – 157 = 594.
That hasn’t quite exposed all of the possibilities. Maybe 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 to start with.
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 was the one single digit number, zero. So that seems as reasonable a conjecture as any at the moment.
Conjecture 4: In Problem 3, the answer is always a multiple of 99.
If you look a little closer at the answers we have been getting, you can see a more precise conjecture.
Conjecture 5: In Problem 3, if the 3-digit number is abc, then the answer is always (a – c) x 99.
Are these conjectures amenable to Proof by Exhaustion? There are certainly less than 900 numbers that have to be dealt with but how many 3-digit numbers abc are there with a ≥ c? From Session 1 we know that there are 54 numbers with a ≥ c. Each of these now has 10 numbers that can be inserted for b. So there are 540 numbers that have to be dealt with here. That seems an awful lot even for a class of 30 to tackle. But does the value of b matter? Isn’t the answer to 543 – 345, the same as 563 – 365? If we can show that, then we are back to 54 numbers and the class might have a show at tackling that in 10 minutes or so.
So what is 5b3 – 3b5? It’s 500 + 10b + 3 – 300 – 10b – 5 = 198. And 198 doesn’t have a b in it. So we can do 10 3-digit numbers in one go!
Now we’ve got that straight, you might guess what the next problem is going to be. Of course it’s the 3-digit version of Problem 2.
Problem 4: Take any 3-digit number; reverse the digits; subtract the smaller number from the larger number to give the number M. Now reverse the digits of M to give the number N. Add M and N. What answer do you get?
Let’s do it for the four examples we have above.
Again we start with experiments.
563 – 365 = 198 and 198 + 891 = 1089; 815 – 518 = 297 and 297 + 792 = 1089.
That looks ominous! Do we always get 1089?
Conjecture 6: In Problem 4, no matter what the original 3-digit number is the answer is always 1089.
It’s actually not difficult to find a counter-example to this though. Start with a palindrome and you will always get zero. So we need to revise the conjecture to give the next conjecture.
Conjecture 7: In Problem 4, no matter what the original 3-digit number is the answer is either 1089 or zero.
Can we prove this by exhaustion? Are there a reasonable number of numbers to tackle? Well if we’ve proved Conjecture 5, we know that we only have to worry about multiples of 99. Between 0 and 900 (the biggest answer to Problem 3 that we can get is 9b0 – 0b9 = 9 x 99 = 891), there are only 10 multiples of 99. So this can be proved by exhaustion too.
Here we look for generalisations of Problems 1 and 2 and see what progress we can make.
Following on from the previous problems the obvious next step is to go to 4-digit numbers and after that 5-digit numbers and after that 6-digit numbers …
Problem 5: Take any 4-digit number; reverse the digits; subtract the smaller number from the larger number. What can you say about the result?
Thinking about the pattern that seems to be forming with Problems 1 and 3 we might be tempted to make the following conjecture
Conjecture 8: If we do the reversing business to 4-digit numbers the answer is always divisible by 999.
However, an example quickly shows this to be false.
4321 – 1234 = 3087 = 3343 x 9. But 3343 isn’t divisible by 111!
Things are starting to fall apart for us here. Not only is Conjecture 8 false but we can’t get a nice (a – b) or (a – c) factor either. The best we appear to be able to do is to get a factor of 9.
Conjecture 9: In the reversing and subtract problem for a 4-digit number, the answer is a multiple of 9.
Can this be established quickly with Proof by Exhaustion? How many cases are there?
Problem 6: Take any 4-digit number; reverse the digits; subtract the smaller number from the larger number to give the number M. Now reverse the digits of M to give the number N. Add M and N. What answer do you get?
Is there any pattern here at all?
Conjecture 10: The answer to Problem 6 is either zero or 10890.
(This conjecture is actually false. To find the correct one, look for Conjecture 10´ in Proof, Level 6.)
Clearly looking at the last two problems shows that we are at the tip on an iceberg. What we would really like to do is to solve the following problems that work for any number of digits.
Problem 7: Take any n-digit number; reverse the digits; subtract the smaller number from the larger number. What can you say about the result?
Problem 8: Take any n-digit number; reverse the digits; subtract the smaller number from the larger number to give the number M. Now reverse the digits of M to give the number N. Add M and N. What answer do you get?
Working on Problem 7 suggests the next conjecture.
Conjecture 11: In the reversing and subtract problem for an n-digit number, the answer is always a multiple of 9.
Can that be proved by Proof by Exhaustion? No. Because here there are an infinite number of cases we cannot check out all of the cases in the list. An algebraic proof can be found in Proof, Level 6, Teachers’ Notes, Session 3.
So we don’t have closure to Problem 7. But what about Problem 8? It would seem to hard at the moment so suppose that n = 4. We’d like to conjecture the following. What do you think? Is the conjecture true or false? If you can find a counter-example, then you will need to invent another conjecture.
Conjecture 12: Suppose that the 4-digit number is abcd, with a ≥ b. Then the answer to Problem 8 is either zero, if abcd is a palindrome; or 10890, if b > c; or 9999 otherwise.
So what about n = 5, 6, …?
What could we do next?
Encourage students to think about what extensions and generalisations they could make. Lead them to the n-digit versions of Problems 1 and 2. Send them off into their groups to work on the extensions to 4-, 5- and n-digit numbers. They should realise by now that they ought to be thinking about experimenting and conjecturing and proving.
Bring the class together to discuss the students’ results.
What conjectures do you have?
Can they be proved by exhaustion?
Is there always a factor of 9 in there?
In this and the next session we extend the problems of the last sessions in another direction.
We seem to have come to an end with the sequence of problems above. But we can try another twist here – addition.
Problem 9: Take any 2-digit number; reverse the digits; add both numbers together. What can you say about the result?
The methodology has been set down in the previous three sessions. Experiment (does your experimentation lead to multiples of 11?); look for conjectures; find proofs or counter-examples. Write things up. Extend or generalise.
In the same way as before try to set up a Proof by Exhaustion.
So now extend to 3-digit numbers. Nothing worthwhile appears to happen. But don’t give up. Push on to 4-digit numbers. Suddenly, things are interesting again. The multiple of 11 reappears but the exact multiple is not obvious. It’s a bit like the situation in Problem 5.
Then you would want to try 5-digit numbers but there seems to be nothing interesting there. But 6-digit numbers are a different thing again. Can you start to see that the n-digit version of Problem 9 gives an interesting result if and only if n is even? This suggests:
Conjecture 13: Let n = 2m and let A be an n-digit number. Reverse the digits of A to form the n-digit number B. Add A and B to form C. Then C is divisible by 11.
The general proof (see Proof, Level 6, Teachers’ Notes, Session 4) may be too hard for most students but you might think about a Proof by Exhaustion for n = 6.
Talk about the way mathematicians like to extend problems. There are lots of examples of this in the Problem Solving section of the site.
How have we extended the problems that we have been working on?
What variations might we try now? (Hopefully they will be able to think of adding instead of subtracting. They may even have thought of this in a previous session. If so, this is the time to recall that.)
Let them work in groups on Problem 9.
Is there a pattern here?
As you move round the groups encourage them to try several 2-digit numbers to get a conjecture.
Can you prove your conjectures?
Here we complete the problem of the last session and look at the additive extension of Problem 2.
The final thing that might be done here is
Problem 10: Take a 2-digit number; reverse the digits; add these two numbers together; reverse the digits of the combined number; add the last two numbers together. Is there any pattern?
Generalise to any number of digits.
Are there any nice patterns here? We’d be interested to hear from you about what you find.
What can we do now?
What extensions can we make to the problems that we have been considering?
Introduce Problem 10 if they are unable to think of it for themselves.
Dear families and whānau,
Recently we have been investigating Conjectures and using Proof by Exhaustion to solve problems involving patterns. Ask your child to share with you one of the problems we have solved as a class, and the the solution they used to solve it.
Printed from https://nzmaths.co.nz/resource/discovery at 9:33pm on the 29th March 2024