Summing Consecutive Numbers : nrich.maths.org Find the number which has 8 divisors, such that the product of the divisors is 331776. Summing Consecutive Numbers Stage: 3 Challenge Level: Swaathi, from Garden International School, started by listing the numbers up to 15 and trying to represent them as sums of consecutive numbers: 2 9 = 4+5 = 2+3+4 10 = 1+2+3+4 11 = 5+6 14 = 2+3+4+5 15 = 7+8 = 4+5+6 = 1+2+3+4+5 We can't write every number as a sum of consecutive numbers - for example, 2, 4 and 8 can't be written as sums of consecutive numbers. In the above, 9 and 15 were the only numbers that I could find that could be written in more than one way. Many people spotted the pattern that all odd numbers (except 1) could be written as the sum of two consecutive numbers.

For example, Matilda and Tamaris wrote: If you add two consecutive numbers together, the sum is an odd number, e.g. 1+2=3 6+7=13 and so on... Well done to pupils from Kenmont Primary School who noticed this, and explained that an Odd plus an Even is always Odd. Some spotted a similar pattern for multiples of 3. Julia and Lizzie said: If you add any 3 consecutive numbers together it will always equal a multiple of 3, e.g. 1+2+3=6 5+6+7=18 Continuing with the patterns, the Lumen Christi grade 5/6 maths extension program team sent us: We discovered that the sum of four consecutive numbers gave us the number sequence 10, 14, 18, 22, 26, 30, and so on. They were all even numbers that had an odd number as half of its total. 1+2+3+4=10 2+3+4+5=14 3+4+5+6=18...

Heather from Wallington High School for Girls explained this pattern: 10 - 1+2+3+4 14 - 2+3+4+5 18 - 3+4+5+6 22 - 4+5+6+7 In all the columns, each place adds 1 each time, so in total you add 4 each time. Ruby said: Numbers which are multiples of 5, starting with 15, are sums of 5 consecutive numbers: 1+2+3+4+5=15 2+3+4+5+6=20 3+4+5+6+7=25... Fergus and Sami noticed a similar pattern: If you allow negative numbers, you can find a sum for any multiple of 7 easily. Each time you add one number either side of the sum, your sum increases by 7, e.g.

3+4=7 2+3+4+5=14 1+2+3+4+5+6=21 0+1+2+3+4+5+6+7=28 -1+0+1+2+3+4+5+6+7+8=35... Great! (There's a way to make this pattern work even without using negative numbers - can you spot it?) Why are all these patterns arising? Becky spotted a different type of pattern: We found out that powers of 2 (2, 4, 8, 16...) can never be made by adding together consecutive numbers together. Interesting! I wonder why? The Lumen Christi team give a way of constructing lots of multiples of odd numbers: We worked out that if you divide a multiple of 3 by 3, and call the answer n, then your original number is the sum of (n-1), n and (n+1). Then we discovered that the multiples of 5 can be written as 5 consecutive numbers. It's the same as the rule for 3 consecutive numbers.

Take a number and divide it by 5, call it n, and then your number is the sum of (n-2), (n-1), n, (n+1) and (n+2). We then made a conjecture that since it is true for 3 and 5, it would also work for 7, 9 and any other odd number. We tested it, and it worked. For example, 63 is a multiple of 7 and 9: 7 numbers: 6+7+8+9+10+11+12=63 (63/7 = 9) 9 numbers: 3+4+5+6+7+8+9+10+11=63 (63/9 = 7) How could we take this investigation further? Arthur asked: Are there any other patterns? Can we explore the powers of two further? Is there a nice way to write certain numbers (for example, every other even number) as a sum of consecutive numbers? Ottilie suggested: Instead of adding, you could multiply the consecutive numbers, and see what patterns come up.

You could also only add consecutive even numbers, or only consecutive odd numbers. These things could all have something in common, or there could be a pattern between them, or nothing at all, maybe? Magnus asked: Is the rule that the powers of two can never be made always true? Can all numbers except the powers of two be made? Great questions! By the way, Abhi sent us a nice algebraic proof that powers of 2 can never be made: Case 1: can we make $2^n$ from an odd number of consecutive numbers? An odd number of consecutive numbers has a whole number as an average. This average is always the middle number. So, that means that the sum of the numbers will be: Sum = average $\times$ number of consecutive numbers. = whole number $\times$ odd number This means the sum has an odd number as a factor. But $2^n$ cannot have an odd number as a factor. This proves that an odd number of consecutive numbers cannot add to make $2^n$. Case 2: can we make $2^n$ from an even number of consecutive numbers? An even number of consecutive numbers will not have a whole number as an average.

The average will be the average of the two middle numbers. So: Sum = (sum of two middle numbers) $\times \frac{1}{2} \times$ number of consecutive numbers = (sum of two consecutive numbers) $\times$ ($\frac{1}{2}\times$ Even number) = (sum of two consecutive numbers) $\times$ whole number But if you add two consecutive numbers, the answer is always an odd number. So a sum like this must have an odd number as a factor again - but $2^n$ doesn't. This proves that an even number of consecutive numbers cannot add to make $2^n$. Nicely done! Consecutive Numbers | nzmaths Consecutive Numbers Purpose: 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 learn what generalisations, extensions, conjectures, theorems, and proofs are work through a completely novel situation and try to develop a mathematical theory around it Achievement Objectives: 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 Description of mathematics: 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. Required Resource Materials: 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 1-stringed if the number can only be written as the sum of one string of consecutive numbers; 2-stringed if it can be written as the sum of two strings; 3-stringed 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 1-stringed; 9 and 25 are 2-stringed; 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 1-stringed; which are 2-stringed; 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. There’s not much here to get you much ammunition on 3-stringed 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 1-stringed. Conjecture 4: All numbers that are the product of a power of 2 and a prime are 1-stringed. Conjecture 5: All numbers that are squares but not powers of 2 are 2-stringed. Conjecture 6: All numbers that are the product of a power of 2 and a square but not powers of 2 are 2-stringed. Conjecture 8: All numbers that are a product of two factors are 3-stringed. Conjecture 9: The only numbers that are stringless are powers of 2.

Conjecture 10: The only numbers that are 1-stringed are primes or products of a power of 2 and a prime. Conjecture 11: The only numbers that are 2-stringed 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 3-stringed 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 counter-example 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 1-stringed, 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 counter-examples. For instance, if they say that the only numbers that are 1-stringed are prime numbers, then someone should be able to point out that 6 is 1-stringed. 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 1-stringed? 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 1-stringed. 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 counter-examples.) 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 over-achiever Gauss knocked it off in no time flat. (To see more about Gauss and other mathematicians of note try the web site www-history.mcs.st-andrews.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 3-stringed. 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 i-stringed 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 2-stringed. 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 1-stringed; and show that 2rp, where r is any number and p is an odd prime, is also 1-stringed. 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, 2r, is the sum of some string.

Then k[2a + k – 1]/2 = 2r or k[2a + k – 1] = 2r+1. Now we know that k is not 1. So, since k must be a factor of 2r+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 1-stringed.

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 2rp. Taking the usual approach we get: k[2a + k – 1]/2 = 2rp, so k[2a + k – 1] = 2r+1p. 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 = 2r+1 and 2a + k – 1 = p. Then 2a + 2r+1 – 1 = p or 2a = p + 1 – 2r+1. So a = (p + 1)/2 – 2r. Case 2: k = p and 2a + k – 1 = 2r+1.

Then 2a + p – 1 = 2r+1or 2a = 2r+1 – p + 1. So a = 2r – (p – 1)/2. We were hoping to show that 2rp was 1-stringed but we seem to have shown that it is 2-stringed. So think a bit. The power 2r+1 is either less than p or bigger than p. They can’t be equal since p is odd. If 2r+1 is less than p we get Case 1; if it’s bigger than p, then we get case 2. So 2rp is indeed 1-stringed. 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.) Add this resource to one of your plans: SOLUTION: find three consecutive odd numbers whose sum is 303 SOLUTION: find three consecutive odd numbers whose sum is 303 You can put this solution on YOUR website!

Consider a series of odd numbers such as 1, 3, 5, 7, 9, 11 and so on. This series consists of consecutive odd numbers because no odd number is left out. Now, notice the particular feature of this series is that each number is 2 greater than the number that comes just before it. So if 5 is the first odd number you select, the next consecutive odd number is 5 + 2 or 7. And think about this. The next consecutive number after that is two more than its predecessor, so it is 7 + 2 or 9. But 7 is equal to 5 + 2. So we can say that the third number in this series of consecutive odd numbers is (5 + 2) + 2 = 5 + 4 = 9. In fact we can write the series of consecutive odd numbers starting with 5 as (5), (5 + 2), (5 + 4), (5 + 6), (5 + 8), and so on. .

In general we can similarly write the series of odd numbers in which n represents the unknown first number, as n, n+2, n+4, n+6, n+8, ... . So three consecutive odd numbers can be written as n, n+2, and n+4. The problem asks you to find three consecutive odd numbers whose sum is 303. You can set up the equation for the sum of three consecutive odd numbers as equal to 303 by writing it as follows: . n + (n+2) + (n+4) = 303 . You can now simplify the left side by adding the three n terms to get 3n and then adding the 2 and the 4 to get 6. This makes the equation become: . 3n + 6 = 303 . First get rid of the 6 on the left side by subtracting 6 from both sides to simplify the equation to: . 3n = 297 . Solve for n by dividing both sides of the equation by 3 (the multiplier of x) to reduce the equation to: . n = 99 . We identified n as the first term in the series.

So the second consecutive odd term must be 2 more that 99, or it must be 101. And the third consecutive odd term must be 2 more than the second term, so it must be 103. . Check the answer by adding 99 + 101 + 103 to ensure that the sum of these three consecutive odd numbers is 303, just as the problem requires it to be. . In summary, the 3 consecutive odd numbers that add to a total of 303 are 99, 101, and 103. . I hope this helps you to understand how to work problems having consecutive odd numbers. . Consecutive Numbers Year 10 Interactive Maths - Second Edition Consecutive Numbers If x is any number, then x and x + 1 are consecutive numbers. E.g. 14 and 15 are consecutive numbers, as are 21 and 22. Example 18 If the sum of two consecutive numbers is 159, find the numbers. Solution: If x is any even number, then x and x + 2 are consecutive even numbers. E.g.

8 and 10 are consecutive even numbers, as are 24 and 26. Example 19 If the sum of two consecutive even numbers is 194, find the numbers. Solution: If x is any odd number, then x and x + 2 are consecutive odd numbers. E.g. 7 and 9 are consecutive odd numbers, as are 31 and 33. Example 20 If the sum of two consecutive odd numbers is 228, find the numbers. Solution: Copyright ï¿½ 2000-2017 mathsteacher.com Pty Ltd. All rights reserved. Australian Business Number 53 056 217 611 Please read the Terms and Conditions of Use of this Website and our Privacy and Other Policies . If you experience difficulties when using this Website, tell us through the feedback form or by phoning the contact telephone number. Quandaries & Queries at Math Central From Lucy: Hello, I have a question about consecutive numbers. So lets say a question tells you to find what three consecutive numbers are that add up to a certain number.

To calculate this, would you use 2x+1, 2x+3, 2x+5 or just x, x+2, x+4. Also, the same question for even numbers. Would you use: 2x, 2x+2, 2x+4 or just x, x+2, x+4. My math book is confusing because it uses the first strategy(the 2x+ï¿½) in some situations and the second strategy(x+ï¿½) in other situations. Thus, can you provide me an explanation of why you would use the strategy that is correct? Specifically, my book answered this question: If ï¿½nï¿½ is the first of five consecutive odd numbers, what is their average, using n,n+2ï¿½.=5n+20 and average is n+4(this is the answer),. For the question: the sum of 4 consecutive odd integers must be: Divisible by 8 but not necessarily 16(this is the answer), it used the strategy 2k+1, 2k+3ï¿½=8k+16.

Thank you so much for taking the time to answer my question. Answered by Harley Weston. 2010-10-21 From Mickey: Find four consecutive integers such that the sum of the first three integers is 20 more then the fourth integer. I know that for consecutive integers the algebraic expression is x+x+1+x+2+x+3 Answered by Penny Nom. 2010-06-28 From Sharon: the product of 2 consecutive integers is -6 less than the square of the larger number. find the smaller number Answered by Penny Nom. the sum of two consecutive integers is 51 Answered by Penny Nom. 2007-09-28 From Joshua: We were given a question that involved consecutive integers. I can not seem to solve this question and I am the top student in my class.The question states: Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91. How do you solve this question? Thank you very much for your support. Answered by Penny Nom, Stephen la Rocque and Claude Tardif.

2007-04-26 From nana: some numbers can be written as the sum of two or more consecutive integers(we consider positive integers only) other number (eg. 4) cannot be express in this way. let us call a number which can be expressed in this way a SOALTCI( sum of at least two consecutive integers) a). list the first few SOALTCI and conjecture the general formula. b). Prove that any number of the given form(in your answer (a)) is a SOALTCI Answered by Penny Nom. 2007-01-08 From Evan: Twice the smaller of two consecutive integers is the larger increased by 5. find the integers. Answered by Stephen La Rocque and Penny Nom. 2006-10-29 From Matias: if the sum of 2000 consecutive integers is 1000, then the sum digits of the greatest of these 2000 integers is? Answered by Penny Nom.

2006-10-29 From Jimmy: Basically i want to know how you prove that the product of any 3 consecutive integers is a multiple of 6 Answered by Stephen La Rocque. 2006-03-12 From Sandy: When I divided the product of three consecutive positive integers by their mean, I got 99. What is the smallest of the three numbers? Answered by Stephen La Rocque. Properties of Consecutive Numbers in Algebra By Deb Russell Updated September 18, 2016. Algebra problems often ask about properties of consecutive numbers, such as 1, 2, 3 or 9, 10, 11. The numbers 3, 6, 9 are not consecutive numbers, but are consecutive multiples of 3. Basically it means that the numbers are adjacent integers. A question may also ask about consecutive even numbers or consecutive odd numbers. These are numbers like 2, 4, 6, 8, 10 or 13, 15, 17. We take one odd number, then the very next odd number after, etc. So, how do we represent these numbers algebraically?

If the question calls for consecutive numbers, I can let one of these numbers be x. Then, the next consecutive numbers will be x + 1, x + 2, etc. I f the question calls for consecutive even numbers, we would have to “make sure” that the number we chose was even. We can do this by letting the first number be 2x instead of x. What would the next consecutive even number be? Be careful. 2x + 1 is not even. So, our next numbers would be 2x + 2 + 2x + 4. Similarly, consecutive odd numbers would be of the form 2x + 1, 2x + 3, etc. continue reading below our video 4 Tips for Improving Test Performance Examples: The sum of two consecutive numbers is 13, what are the numbers? Let the first number be x and the second number be x + 1. Then: x = 6 So, our numbers are 6 and 7. What if we had chosen our consecutive numbers differently from the start? Would we get a different answer?

Let’s see: Let the first number be x - 3, and the second number be x - 4. These numbers are still consecutive numbers: one comes directly after the other. (x - 3) + (x - 4) = 13 2x - 7 = 13 2x = 20 x = 10 We have x equal to 10 here and we had x equal to 6 above!! What’s going on? Well, substitute 10 x = into the variables we chose and see what happens. 10 - 3 = 7 10 - 4 = 6 We get the same answer. Everything works out. Sometimes it makes things easier if you choose different variables for your consecutive numbers. For example, if you had a problem involving the product of 5 consecutive numbers, which would you rather calculate?

x (x + 1) (x + 2) (x + 3) (x + 4) or (x - 2) ( x - 1) (x) (x + 1) (x + 2) The second equation is a lot more friendly because it can take advantage of properties of the difference of squares! Try Some On Your Own Try these consecutive number problems on your own. Even if you can figure some of them out without the methods above, try them using consecutive variables for practice: 1. 4 consecutive even numbers have a sum of 92. What are the numbers? 2. Five consecutive numbers have a sum of zero. What are the numbers? 3. Two consecutive odd numbers have a product of 35. What are the numbers? 4. Three consecutive multiples of five have a sum of 75. What are the numbers? 5. The product of two consecutive numbers is 12. What are the numbers? 6. The sum of 4 consecutive integers is 46. What are the numbers? 7. The sum of 5 consecutive even integers is 50. What are the numbers? 8.

If you subtract from the product of 2 consecutive numbers, the sum of the same two numbers, the answer is 5. What are the numbers? 9. Do there exist 2 consecutive odd numbers with product 52? 10. Do there exist 7 consecutive integers with sum of 130? Solutions 1. 20, 22, 24, 26 2. -2, -1, 0, 1, 2 3. 5, 7 4 Ways to Add 5 Consecutive Numbers Quickly - wikiHow Using the Middle Number 1 Mentally multiply the middle number by 5 ... finished!? That's all you do!. e.g. with your example 53 X 5 = 265. Here's how to multiply this mentally: First separate 53 into 50 and 3. Now 50 X 5 = 250. Also 3 X 5 = 15. Now add those separate answers. 250 + 15 = 265. 2 Learn how it works: Let the smallest number be (x - 2).

Then the other 4 are (x - 1), (x), (x + 1) and (x + 2). The sum: (x - 2) + (x - 1) + (x) + (x + 1) + (x + 2) = 5x Using the above method: 10x/2 = 5x Method If this question (or a similar one) is answered twice in this section, please click here to let us know. Tips You can add any sequence of consecutive numbers, even or odd, regardless of the number of integers the sequence contains, by adding the first and the last number in the sequence, dividing it by two and multiplying the answer by the number of integers in the sequence. In algebra , we can say ((a+b)/2)*n, or rearrange it to n*(a+b)/2 to remove a set of parenthesis. Method 2 can be used for any odd amount of consecutive numbers, but instead of using "5x" you must use "(amount of consecutive numbers)x" e.g. in 6 + 7 + 8, seven is x.

(3)7 = 21, and 6 + 7 + 8 = 21 Advanced Usage They don't need to be consecutive numbers. They just need to be a sequential subset of any linear equation. ( The examples above use the linear equation of x = c + 1 * n ) For example, let's use the linear equation x = 10 + 7y, therefore, {xϵN| 17,24,31,38,45,...} So if we use: 17, 24, 31, 38, 45 31 x 10 = 310 and 310/2 = 155 They also do not need to be integers.

Let's use the linear equation x = 1 + y / 20, therefore, {xϵN| 1.05, 1.1, 1.15, 1.2, 1.25, ...} So if we use: 1.05, 1.1, 1.15, 1.2, 1.25 1.15 x 10 = 11.5 and 11.5/2 = 5.75 They also do not need to be positive values. The set can contain negative numbers, positive numbers, or both. This method can be extended to be used (as above) for any ODD number of consecutive integers --> 5, 7, 13, 25, 99 consecutive as long as you can locate the median digit and then multiply it be the number of integers. (Example 12, 13, 14, 15, 16, 17, 18, 19, 20 = 144 = 16(median) x 9(integer count).

This can be most impressive if coupled with the simple trick of multiplying by 11 . Video

Question: If 3 consecutive numbers add up to 27, what is the smallest of those numbers?
Answer: 8