Here we look for patterns in some number problems and see how far we can extend two basic ideas. The main point of this unit is to produce conjectures although we will spend a little time on proofs.
 work through the steps required of a complicated problem
 investigate special cases of a problem
 make conjectures
 provide simple justifications
This unit illustrates the way that pure mathematics works from posing a problem to experimenting to conjecturing to proving to extending and generalising. So it employs general strategies that can be used time and time again. Although we use these strategies here 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 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 means that we can prove something by looking at every case. This method is covered in detail in the Teachers’ Notes in the early sessions of the unit.
This unit only requires paper and pencil.
Session 1
In this session we look at the basic two problems of this unit and see what progress can be made with them.
Teachers’ Notes
Problem 1: Take any 2digit 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 2digit number gives us a multiple of 9.
Conjecture 1: If we take any 2digit 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 2digit number.
It’s worth noting that Conjecture 2 includes multiples of zero.
Problem 2: Take any 2digit 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 (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.
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 only going to be two answers, 0 and 99.
Conjecture 3: In Problem 2, the answer will either be 99 or 0.
The conjectures we have so far can all be proved by Exhaustion. That is by exhausting all cases. We can prove Conjectures 1 and 2 by looking at all of the 90 2digit numbers to see what answer they give.
Actually, when you think about it, it’s not that many. After all we only have to worry about the 2digit 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 by Gauss’method – as described in the Background to Algebra 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!
And 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.
Teaching Sequence
 Introduce Problem 1 to the whole class. Discuss how you might go about solving it. Encourage them to draw up a table on the board. Fill in a few entries on the table (see Teachers’ Notes).
 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.)
 Go from group to group offering help and encouragement.
 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 them 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 counterexamples so they are only probably true. Encourage the more able students to think 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 2digit 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 2digit 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?
 Now consider Problem 2. After an initial discussion to ensure that everyone understands the problem, let them work in their groups on finding conjectures.
 Go from group to group offering help and encouragement.
 Let different groups report on their findings to the whole class.
Discuss what they report.
Be sure to make them aware that we don’t know if these conjectures are true or not. All that we can say at this point is that we can find no counterexamples.
Encourage the more able students to think how you might prove the conjectures that have been made. Proof by Exhaustion is one approach.
Is that feasible here?
How many calculations will be needed?
How many 2digit numbers are there?
Do we have to test all of these numbers?
But is this relevant? Why don’t we only need to look at multiples of 9?
Is it feasible for us to divide these up between us?
 Let them write their conjectures and an outline of any proofs in their books.
Session 2
In this session we extend Problems 1 and 2 and try to make conjectures about their answers.
Teaching Notes
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 3digit 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 2digit answer. How could we do that? 564 – 465 = 99. That’s one possibility. And could we get a 1digit 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 3digit 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 2digit 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.
Actually if you look a little closer at the answers we have been getting you can actually see a more precise conjecture than this.
Conjecture 5: In Problem 3, if the 3digit 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 3digit 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 40 to have 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 3digit 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 3digit version of Problem 2.
Problem 4: Take any 3digit 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 3digit number is the answer is always 1089.
It’s actually not difficult to find a counterexample 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 3digit 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 knocked off by exhaustion too.
Teaching Sequence
 Let them recall what was done in Session 1.
What were the problems we looked at?
What conjectures did we make?
How did we use Proof by Exhaustion?
What Problems could we make up that look something like Problems 1 and 2? (Move them towards Problems 3 and 4. If they think up some of the later problems of this unit, tactfully put them on hold till a later session.)
 The approach now is essentially the same as in Session 1. The only real difference will be when the students are considering proving the conjectures by exhaustion. You will then need to lead them through the counts that help them to realise that the numbers that they have to deal with are small.
 Give the students time to write up the conjectures in their books along with a hint (two or three examples) of how they were proved.
Session 3
Here we look for generalisations of Problems 1 and 2 and see what progress we can make.
Teachers’ Notes
Following on from the previous problems the obvious next step is to go to 4digit numbers and after that 5digit numbers and after that 6digit numbers …
Problem 5: Take any 4digit 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 4digit 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 4digit number, the answer is a multiple of 9.
Can this be established quickly by Proof by Exhaustion? How many cases are there?
Problem 6: Take any 4digit 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 ndigit number; reverse the digits; subtract the smaller number from the larger number. What can you say about the result?
Problem 8: Take any ndigit 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 ndigit 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 counterexample, then you will need to invent another conjecture.
Conjecture 12: Suppose that the 4digit 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, …?
Teaching Sequence
 With the whole class together, get them to recall what has been done over the last two sessions. (Problems 1 to 4 with 2 and 3digit numbers along with their conjectures and proofs.)
 What could we do next?
Encourage them to think about what extensions and generalisations they could make. Lead them to the ndigit versions of Problems 1 and 2. Send them off into their groups to work on the extensions to 4, 5 and ndigit numbers. They should realise by now that they ought to be thinking about experimenting and conjecturing and proving.
 Help the various groups as you move around the class. Get them to think about what conjectures could be made and how proofs might be established.
 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?
 On the board, with students’ help, try to write up a proof of the answer to Problem 5. Get them to write the problem and its solution in their books.
 What do they have to say about the ndigit case? (Problem 7.)
 Then repeat the steps above with Problem 6 and its generalisation Problem 8.
Session 4
In this and the next session we extend the problems of the last sessions in another direction.
Teachers’ Notes
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 2digit 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 counterexamples. Write things up. Extend or generalise.
In the same way as before try to set up a Proof by Exhaustion.
So now extend to 3digit numbers. Nothing worthwhile appears to happen. But don’t give up. Push on to 4digit numbers. There 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 5digit numbers but there seems to be nothing interesting there. But 6digit numbers are a different thing again. Can you start to see that the ndigit 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 ndigit number. Reverse the digits of A to form the ndigit 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.
Teaching Sequence
 Remind the students of the problems that we have done in the last three sessions. Discuss how they went about collecting evidence for the conjectures and how they finally solved the problems. You might want to recall how they worked on special cases and how these special cases led to conjectures and how these conjectures were proved.
 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 2digit numbers to get a conjecture.
Can you prove your conjectures?
 Have a reporting back session. Discuss the work to date and try to settle Problem 9. Give them time to write up that part of the work.
 Have a discussion as to what you might do next. They should think about the 3digit and 4digit cases of Problem 9.
 Let them go into groups to work on the problems that come from extending Problem 9. Go around the groups to provide help and ask leading questions.
 There is quite a bit of work here so it may spill over into the next session.
Session 5
Here we complete the problem of the last session and look at the additive extension of Problem 2.
Teachers’ Notes
The final thing that might be done here is
Problem 10: Take a 2digit 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.
Teaching Sequence
 Complete the work of the last session.
 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.
 Now work as we have suggested in the last few sessions. Let the students do most of the work in groups but get them to report what they have found in a class discussion. Finally ask them to present their work.