This problem solving unit is suitable for Level 5 (or Level 6) students.
In this problem solving unit, we look at the way numbers can be written as the sum of consecutive strings of whole numbers. The point of this unit is to give students a chance to:
 see how mathematicians operate
 display ingenuity and creativity
 practice arithmetic in context
 learn what generalisations, extensions, conjectures, theorems, and proofs are
 work through a completely novel situation and try to develop a mathematical theory around it
solve a mathematical problem
see how to generalise and extend a problem
understand why and how mathematical statements may be justified
work with others to solve a problem and generate ideas
Like all of the Problem Solving units, this one aims to introduce students to the underlying ideas of mathematics through a problem. The problem here requires a knowledge of arithmetic and the use of some algebra. In this unit we see how a mathematical theory might develop through experimentation, conjecturing, proving, generalising and extending.
As with all of these units it is difficult to break the development here up into lessons as we can never be quite sure how any particular class will progress. This will depend both on their ability and on your scaffolding. The right question asked at the right time will enable more rapid progress.However, you don’t want to make it too rapid! Most of the class should be able to follow what you are doing.
This unit is potentially a very long one. You might want to split it into two or even focus it on one or two aspects such as which numbers can’t be expressed as the sum of a consecutive string of whole numbers or which ones can be so expressed in only one way. It might be a good idea to get many of the concepts of this unit out of the way in previous lessons. In particular, you might do a problem solving lesson that uses the ideas of experimentation and conjecturing; you might introduce a formula for the sum of any string of consecutive whole numbers; and you might do an example of a proof by contradiction. At any point that you feel that you are getting bogged down, go back to an example or examples to try to get the general feel for what is going on.
Lesson Sequence
Session 1
In the first session of this unit, the class work on a large range of numbers to determine when they can be written as the sum of a string of consecutive whole numbers.
Background
First we start off with the problem.
Problem 1: Is it possible to find a string of consecutive whole numbers that add up to 2500?
Let’s start off with a few of observations. First, clearly 9 can be written as the sum 2 + 3 + 4. So it is certainly possible to find a string of consecutive numbers that add up to 9. In fact 4 + 5 = 9, so there is more than one string that adds up to 9.
Second, this problem, in a sense, has nothing to do with 2500. 2500 is just a big number to get the unit moving. Any number will do for that purpose, though probably a big number is more interesting and challenging than a small number.
Third, there are at least two ways to do this problem – with a bit of logic and some arithmetic; and using algebra. In fact if you let the students have open slather on this one, then they may just add a few strings of consecutive numbers together and give you the sum as a number that is the total of a string of consecutive numbers. Now there is no need to let on that they are ‘tricking’ you. The strings that they get will all give some evidence to put together for the later sessions.
Fourth, we’ll assume that a string of numbers has at least two numbers. Obviously there wouldn’t be too much fun here if we let a single number be a string. Then any number would be the sum of a string.
And finally, the problem talks about whole numbers so from now on we won’t mention the fact again. We’ll assume that we are using just the numbers 1, 2, 3, … We won’t ever use negative numbers or zero in a string of consecutive numbers in the whole of this unit.
2500 is too large. So try some smaller numbers and see what your class comes up with. In this first session get them to try as many different numbers as you can so that you all have some ammunition with which to operate in the later sessions.
It might help you to know that the only numbers that cannot be written as the sum of a string of consecutive numbers are the powers of 2. We’ll justify this later on.
Now it might not come as a surprise but the only numbers that can only be expressed as the sum of one string of consecutive numbers are the primes AND the primes multiplied by a power of 2. We’ll justify that too.
The aim of this first session is to look at as many numbers as possible and make up conjectures based on the results. To this end, you might give different numbers to different people in the class. If they all have, say, three numbers to work on you’ll have quite a pile of ammunition to conjecture with in Session 2. Students who finish their numbers early could be allowed to choose any numbers outside the range that the class was working on.
It might help you to have some words handy to enable you to avoid long mouthfuls of ‘strings of consecutive numbers that add up to’type talk. This problem isn’t sufficiently famous to have terms already defined. We’ll (you’ll) have to make them up. So we’ll say here that a number is stringless if it cannot be written as the sum of a string of consecutive numbers and stringed if it can. Later on you might need to have 1stringed if the number can only be written as the sum of one string of consecutive numbers; 2stringed if it can be written as the sum of two strings; 3stringed etc., etc. But your students might get a buzz out of making up their own terms. You don’t have to use the stuffy ones we’ve invented.
With a bit of work you should be able to see that 8 and 32 are stringless; 13 and 34 are 1stringed; 9 and 25 are 2stringed; and so on.
Teaching Sequence
1. Introduce the problem by doing an example or two yourself. Show that some numbers are the sum of more than one string. Then get them to do some small examples as a whole class exercise. When you think most students have the feel for the problem, set them to work in pairs to produce more examples. You might give one group the numbers 11, 21 and 31; another group 12, 22 and 32; and so on. This way you might cover 50 or more numbers.
2. Go round the groups and check their work. They should be trying to find out (i) if the number they are working on can be expressed as a string and (ii) in how many ways it can be expressed as a string. Make sure that they are on track. Give groups that have finished their allotted numbers a chance to choose number of their own to try.
3. After a suitable time, bring the whole class together to report on what they have found. What ideas do they have about the way numbers are behaving?
4. Discuss how they might record the results of their work. Get the students to put the results in their books for later reference.
(You might get more entries in your table if you get the students to do some numbers for homework after Session 1. If you do, get them to think about what is going on and have some ideas ready for the next session.)
Session 2
In this session students make conjectures about which numbers are stringless, which are stringed, which are 1stringed; which are 2stringed; and so on. The conjectures are tested.
Background
This is the conjecture session. Here you should be looking for patterns. Perhaps the best way to do that is to draw up a table on which the students can enter their results. Then they’ll have evidence to make conjectures from. Below we have started a table for you.
No. 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
17 
18 
19 
stringless 
√ 
√ 

√ 



√ 







√ 



stringed 


√ 

√ 
√ 
√ 

√ 
√ 
√ 
√ 
√ 
√ 
√ 

√ 
√ 
√ 
1stringed 


√ 

√ 
√ 
√ 


√ 
√ 
√ 
√ 
√ 


√ 

√ 
2stringed 








√ 








√ 

3stringed 














√ 




There’s not much here to get you much ammunition on 3stringed numbers but with the whole class at work you should get some worthwhile data. (You might get a lot more entries in your table if you get the students to do some numbers for homework after Session 1.)
Here are some conjectures that the students might come up with. If they don’t, you might give them a gentle prod. The conjectures are numbered in no specific order. Numbering them just enables us to refer to them more easily later on.
Conjecture 1: All odd numbers are stringed.
Conjecture 2: All powers of 2 are stringless.
Conjecture 3: All prime numbers are 1stringed.
Conjecture 4: All numbers that are the product of a power of 2 and a prime are 1stringed.
Conjecture 5: All numbers that are squares but not powers of 2 are 2stringed.
Conjecture 6: All numbers that are the product of a power of 2 and a square but not powers of 2 are 2stringed.
Conjecture 8: All numbers that are a product of two factors are 3stringed.
Conjecture 9: The only numbers that are stringless are powers of 2.
Conjecture 10: The only numbers that are 1stringed are primes or products of a power of 2 and a prime.
Conjecture 11: The only numbers that are 2stringed are squares or products of a power of 2 and a square, provided that they are not powers of 2.
Conjecture 12: The only numbers that are 3stringed have two factors or are products of a power of 2 and a number that has two factors, provided that they are not powers of 2.
Conjecture 13: If a number N is divisible by 3 then there is a string of three numbers that adds up to N.
Conjecture 14: If a number N is divisible by 5 then there is a string of five numbers that adds up to N.
Conjecture 15: If a number N is divisible by 4 then there is a string of four numbers that adds up to N.
Conjecture 16: If a number N is divisible by m then there is a string of m numbers that adds up to N.
Conjecture 17: Any sum of three consecutive numbers adds up to a number that is divisible by three.
Conjecture 18: Any sum of five consecutive numbers adds up to a number that is divisible by five.
Conjecture 19: Any sum of four consecutive numbers adds up to a number that is divisible by four.
Conjecture 20: Any sum of m consecutive numbers adds up to a number that is divisible by m.
Not all of these conjectures are true, though most of those that are false can be reworded to produce a true conjecture.
After having come up with the conjectures the class should spend some time checking them out. Can they find an odd number that is stringless, for instance? If so, Conjecture 1 is false. Can they find a power of 2 that is stringed? If so, Conjecture 2 is false.
Be careful when a student finds a counterexample like this. He or she may have made an error. Get the student to justify their reasoning. Make sure that they can convince everyone else in the class, including you.
Note the subtle difference between Conjecture 2 and Conjecture 9. The first of these says that powers of 2 are stringless. But the second says that there are no other stringless numbers than powers of 2. Conjecture 9 is a much more powerful conjecture than Conjecture 2. If Conjecture 9 is true then Conjecture 2 just has to be.
Spend this session producing conjectures, checking them, and rewording those that you can find counterexamples for. This might take more than one lesson. You could, perhaps, let the students work on one of the conjectures for homework.
Teaching sequence
1. If you haven’t already done so, record the results so far on the board or a large piece of paper that can be seen by everyone in the class.
2. Tell them that in this session they are looking for patterns/conjectures. They can propose any pattern that they see. However, they have to have tested the pattern for several numbers before they can propose it.
3. Before they go away to work in pairs/groups, look at the results that you have and see if there are any conjectures that seem to stand out. Also look to see if there are any errors in the table. You may find that some of these stand out when you look for patterns.
4. Get the students to work in their groups. They have to find at least 3 conjectures. Each conjecture has to be supported by several numbers. For instance, if they say that prime numbers are 1stringed, then they have to be able to show that all of the prime numbers in the range of numbers on the board have only one string.
5. Go round the class and see that the groups are on track. Do not hesitate to give them hints, though try not to tell them any specific conjectures.
6. When you think that they are ready, bring the class together and see what conjectures they have found. Write these so that everyone can see them. As the students give you their conjectures, they should be able to support what they say with many examples of numbers that behave that way. Ask the class if there are any counterexamples. For instance, if they say that the only numbers that are 1stringed are prime numbers, then someone should be able to point out that 6 is 1stringed. In these cases, the conjectures should be adjusted to take the new evidence into account.
7. Feel free to suggest conjectures if the class hasn’t managed to cover all of the ones that you know.
8. Make sure that the class records the conjectures in their books.
(You might let them try to find some conjectures for homework. They could even be asked to think about reasons why their conjectures are true.)
Session 3
In this session, we look at ways to justify the conjectures of the last session. Don’t worry if the justifications are not ‘solid’ proofs. It is enough that they get the idea of the proof.
Background
This session is about trying to find justifications (proofs) for the conjectures that you have been working on in Session 2. We’ll prove a few and give a few counterexamples here. Similar proofs work for similar conjectures. Note: (i) that we’ve numbered the theorems to coincide with the appropriate conjecture. So Theorem 18 comes from Conjecture 18; and (ii) that we put a □ at the end of a proof to show that we have finished.
Theorem 18: Any sum of five consecutive numbers adds up to a number that is divisible by five.
Proof: Let a, b, c, d, and e be five consecutive numbers. Then we can write them more simply in terms of c. This is because a = c – 2, b = c – 1, d = c + 1 and e = c + 2. So a + b + c + d + e = (c – 2) + (c – 1) + c + (c + 1) + (c + 2) = 5 c – 2 – 1 + 1 + 2 = 5c. Hence five consecutive numbers sum to a number that is divisible by 5.
This raises two comments.
Comment 1: This statement almost goes the other way round too. If we have a number that is divisible by 5, then we can express it as the sum of five consecutive numbers – with a few exceptions. For instance, 5 = 1 + 0 + 1 + 2 + 3 and 10 = 0 + 1 + 2 + 3 + 4, don’t fit in with the caveat that we made earlier that we were only dealing with whole numbers here.
But with the exceptions 5 and 10 taken out of consideration the statement of Theorem 18 does go the other way. We just work backwards on the proof of the theorem.
Suppose that n = 5c. Then n = (c – 2) + (c – 1) + c + (c + 1) + (c + 2). So n is the sum of five consecutive numbers provided that c – 2 and c – 1 are positive. This requires c to be at least 3, so n has to be at least 15.
Because of the above we now know that if we have the sum of five consecutive numbers, that sum is divisible by 5 and if we have a number that is divisible by 5, but not 5 or 10, then it can be expressed as the sum of five numbers. We write theorems like that with ‘if and only if’ statements.
Theorem 18: Let n be any number bigger than 10.& Then n is divisible by 5 if and only if it can be expressed as the sum of five consecutive numbers.
Comment 2: Given Comment 1 we should be able to answer Problem 1 from Session 1. Clearly 2500 = 5 x 500. So 2500 can be written as the sum of 498, 499, 500, 501 and 502.
Comment 3: We should also try to generalise this theorem. For what other numbers m is it true that a number is divisible by m if and only if it can be expressed as the sum of m consecutive numbers (Conjecture 20)?
You’ve probably found out by now that this doesn’t hold true for m = 2. And it doesn’t hold for m = 4 either. In fact, it doesn’t hold for any even m. You can see this for m = 4 by looking at the four consecutive numbers a, a + 1, a + 2, and a + 3. They add up to 4a + 6. No matter what a is, 4a + 6 is never divisible by 4 because 6 isn’t.
Let’s move on to some other conjectures and their proofs. Conjecture 1 isn’t quite correct. It clearly falls down for the odd number 1. But with that modification we get Theorem 1.
Theorem 1: All odd numbers bigger than 1 are stringed.
Proof: Any odd number n bigger than 1 can be written in the form n = 2a – 1, where a ≥ 2. From this we can see that n = (a – 1) + a. So n is stringed unless for some reason a – 1 ≥ 1. But a ≥ 2, so a – 1 ≥ 1 and we have both a – 1 and a positive.
Comment 4: This immediately tells us that any stringless number has to be even.
Comment 5: So we know by Theorem 1, that odd prime numbers p, are not stringless but could they be more than 1stringed? Suppose that we have a string of consecutive numbers that has an even number of numbers in it and suppose that it adds up to p. Then for the sum to be odd, it will have to contain an odd number of odd numbers. Suppose then that the string contained 2k numbers, where k was odd. (The string has to have k even and k odd numbers.) By Conjecture 20, the sum of that string would have to be divisible by k. So k is either 1 or the prime p itself. If k = 1, then we have the string a – 1, a from the Proof of Theorem 1. If k = p, then the even length string is made up of two strings of length k. Both of these is a multiple of k by the Proof of Theorem 18. Say they are kc and kd. Then the sum of the total string is kc + kd = k(c + d). Since c = d is bigger than 1, k(c + d) = p(c + d) > p.
So we get:
Theorem 3: Every odd prime is 1stringed.
As the various conjectures are proved to everyone’s satisfaction, then they should be written up in the students’ notebooks.
That ought to be enough proofs for this session.
Teaching Sequence
1. Talk about the need for justification. Say how essential it is in mathematics. Take one of the conjectures that you think they will be able to justify and talk them through it as a whole class activity. Let the students make as many contributions as possible.
2. Let the class know that their task for this session is to try to justify as many conjectures from the last session as they can. Warn them that you will expect them to present some of their work to the whole class. They can choose which conjectures they want to tackle.
3. Set them to work in their groups. Walk around to give assistance. Hear their justifications. Get them to write these down in preparation for presenting them to the whole class.
4. After a while bring the class together and get students to present their work. This needs to be done in a supporting atmosphere where everyone is heard and everyone’s views are treated with respect.
5. The edited work of the groups should be written out carefully and each student given a copy.
(For homework you might photocopy a few of the justifications and ask them to see which of them they think are good justifications and which they think need some improving. You might also get them to check that there are no counterexamples.)
Session 4
In this session we produce a formula for the sum of strings of consecutive numbers. This enables us to find the number of strings that sum to any given number.
Background
In Algebra Unit, The Why and How of General Terms, we show that the sum of the first n whole numbers is n(n + 1)/2. How do we find the sum of the s numbers starting at a? In other words how do we find the sum a + (a + 1) + (a + 2) + … + (a + k – 1)? We’ll do a specific case first.
Let S = 5 + 6 + 7 + 8 + 9 + 10 + 11. Then write this the other way to give
S = 11 + 10 + 9 + 8 + 7 + 6 + 5.
Now add the two lots of S together. We get
2S = 16 + 16 + 16 + 16 + 16 + 16 + 16.
So 2S = 7 x 16 = 112. This means that S = 56.
You should remember that this as the old Gauss trick. The mathematician Gauss was asked by his teacher to add up the first 100 numbers. The teacher had expected that to be a difficult problem. After all it had probably kept many a class before Gauss’ quiet until lunch. Naturally our mathematical overachiever Gauss knocked it off in no time flat. (To see more about Gauss and other mathematicians of note try the web site wwwhistory.mcs.standrews.ac.uk/history.)
In The Why and How of General Terms, we show how to do this by a geometrical method.
Anyway we can now do that in general.
Let S = a + (a + 1) + … + (a + k – 2) + (a + k – 1). Then
S = (a + k – 1) + (a + k – 2) + … + (a + 2) + a.
So 2S = [a + a + k – 1] + [a + 1 + a + k – 2] + … + [a + k – 2 + a + 2] + [a + k 1 + a].
Now this is clearly equal to a multiple of 2a + k – 1. The question is, what multiple? Well it is just the number of numbers that we are adding. There are k of these. So
2S = k[2a + k – 1].
This give S = k[2a + k – 1]/2.
Actually the neat thing about this is that it is the number of terms in the string, multiplied by the sum of the first and the last terms, divided by 2. And it is this that enables us to prove a few things about stringed and stringless numbers.
Let’s see how many strings give us 30. So suppose that
a + (a + 1) + (a + 2) + … + (a + k – 1) = 30.
By what we have just done we know that k[2a + k – 1]/2 = 30.
So k[2a + k – 1] = 60. But the left side must be two factors of 60. So if we list all of the factors of 60 we can equate them to the variables on the left. Look!
First all of the pairs of numbers that can give you 60:
60 = 1 x 60 = 2 x 30 = 3 x 20 = 4 x 15 = 5 x 12 = 6 x 10.
Now let’s equate these to k and 2a + k – 1, bearing in mind that k < 2a + k – 1.
Case 1: k = 1 and 2a + k – 1 = 60. If k = 1, then 2a + 1 – 1 = 60, so a = 30. This gives the string with k = 1 term, and the term is 30. But strings with only one number aren’t allowed in this problem.
Case 2: k = 2 and 2a + k – 1 = 30. Then 2a + 1 = 30. But there is no whole number solution of this.
Case 3: k = 3 and 2a + k – 1 = 20. Then 2a + 2 = 20 or 2a = 18. So a = 9. This gives the string 9, 10, 11.
Case 4: k = 4 and 2a + k – 1 = 15. Then 2a + 3 = 15 or 2a = 12. So a = 6. This gives the string 6, 7, 8, 9.
Case 5: k = 5 and 2a + k – 1 = 12. Then 2a + 4 = 12 or 2a = 8. So a = 4. This gives the string 4, 5, 6, 7, 8.
Case 6: k = 6 and 2a + k – 1 = 10. Then 2a + 5 = 10 or 2a = 5. But there is no whole number solution of this.
So we now have found a systematic way to find the number of strings that make up any number. From the algebra above we see that 30 is 3stringed.
Teaching Sequence
Notes
(1) if the students haven’t met the sum of k whole numbers before, then you won’t do this session in one lesson. In fact it is a good idea to make sure that you do this summation before you start this unit.
(2) In this session, do lots of specific examples. This will enable you to check the istringed properties that you have found so far (by what may have been a hit and miss approach). It will also help you to add to your table from Session 2.
1. Recall the sum of the first n whole numbers and the sum of any k consecutive whole numbers. If you feel it is necessary, you might get the students to sum several strings of numbers using the formula.
2. Show them how to find the number of strings that add up to particular numbers by using the formula for k consecutive numbers.
3. Get then to work in their groups on more examples. Assign specific numbers for them to start on and then let them find their own. You might get them to think about what is the largest number of strings that a number less than 100 can be written as. Or, whether there is a number in the 90s that is 2stringed.
4. Give students who haven’t presented their work before the chance to do so in front of the whole class.
5. Get the students to summarise their work in their books.
(For homework they could be asked to use the method on some more numbers. They might also be asked to see if they can use the method to prove some conjectures such as Conjecture 3.)
Session 5
In this final session we look a bit more deeply into the Conjectures of Session 2. In particular, we see why powers of 2 are stringless; show again that odd primes are 1stringed; and show that 2^{r}p, where r is any number and p is an odd prime, is also 1stringed.
Background
Conjecture 2: Let’s have a look first at Conjecture 2.
We use a proof by contradiction, using the technique of Session 4. Suppose that the power of 2, 2^{r}, is the sum of some string. Then
k[2a + k – 1]/2 = 2^{r} or k[2a + k – 1] = 2^{r+1}.
Now we know that k is not 1. So, since k must be a factor of 2^{r+1}, k must itself be a power of 2.But since k is less than 2a + k – 1, 2a + k – 1 must also be a power of 2. But if k is a power of 2 it is certainly even and k – 1 is odd. So 2a + k – 1 has to be odd too. If 2a + k – 1 is odd it can’t be a power of 2. This contradiction settles the matter.
(If you’re class hasn’t met this technique before it might be a good idea to introduce them to the proof that √2 is irrational first.)
Conjecture 3: We know that we have already settled this but we’ll do it again here by the method of Session 4.
The idea is that k[2a + k – 1]/2 = p, and so k[2a + k – 1] = 2p. As p is an odd prime, the only factors of the right hand side are 1 and 2p or 2 and p. This gives us:
Case 1: k = 1 and 2a + k – 1 = 2p. If k = 1, then 2a + 1 – 1 = 2p, so a = p. This gives the string with k = 1 term, and the term is p. But strings with only one number aren’t allowed in this problem.
Case 2: k = 2 and 2a + k – 1 = p. Then 2a + 1 = p. This gives 2a = p – 1, so a = (p – 1)/2 – a perfectly good whole number. So the string here consists of two terms, (p – 1)/2 and (p + 1)/2.
So you can see that odd primes are 1stringed.
Conjecture 4: Naturally we have to take the product here of a power of 2 and an odd prime. We’ll write this number as 2^{r}p. Taking the usual approach we get:
k[2a + k – 1]/2 = 2^{r}p, so k[2a + k – 1] = 2^{r+1}p. There are lots of potential possibilities here so let’s cut them down at the start.
First we know that we can’t have k = 1. Then we know that if k is even, that 2a + k – 1 has to be odd. And if k is odd, then 2a + k – 1 is even. So we get only two cases.
Case 1: k = 2^{r+1} and 2a + k – 1 = p. Then 2a + 2^{r+1} – 1 = p or 2a = p + 1 – 2^{r+1}. So a = (p + 1)/2 – 2^{r}.
Case 2: k = p and 2a + k – 1 = 2^{r+1}. Then 2a + p – 1 = 2^{r+1}or 2a = 2^{r+1} – p + 1. So a = 2^{r} – (p – 1)/2.
We were hoping to show that 2^{r}p was 1stringed but we seem to have shown that it is 2stringed. So think a bit. The power 2^{r+1} is either less than p or bigger than p. They can’t be equal since p is odd. If 2^{r+1} is less than p we get Case 1; if it’s bigger than p, then we get case 2.
So 2^{r}p is indeed 1stringed.
Teaching Sequence
1. Lead up to each of the three conjectures discussed in the Background above, by doing specific examples. So start off by reminding them how they used the summation formula to show that 16, say was stringless. Work through two or three examples as a whole class. Then work through the general situation. At each step ask them what you should do next and why does that work. Discuss any difficulties that they have.
2. Let them record the general proof in their books.
3. Repeat the approach that you used on Conjecture 2 with Conjecture 3. First discuss particular examples of primes and then work on the general situation. Let them record the work in their books.
4. For Conjecture 4, send them off in their groups to work on specific numbers that are the product of a prime number and a power of 2. Then challenge them to work out the general situation.
5. Go round the groups and give them assistance where necessary. Groups that finish ahead of the others might try some of the other conjectures that relate directly to the number of strings a particular number has.
6. Let students who haven’t presented before show the rest of the class what they have done. Allow collegial discussion.
7. Let the class write up the edited version of the proof.
(For homework let them think about the relation between the factors of a number and the number of ways it can be written as a string. There is still a lot more in this problem that might be used later.)