Neighbourly Addition
Problem
As I walked down the street this morning, I noticed that all of my neighbours' house numbers were odd!
I added three house numbers together as I walked past: 7 + 9 + 11 = 27
Further down the road, I passed some bigger numbers. I added another set of three neighbouring house numbers: 15+17+19 = 51
Can you find some other totals I could make, by adding together the house numbers of three (odd) next-door-neighbours?
Once you've found a few totals, here are some questions you might like to explore:
Is there anything special about all the totals?
Is there a quick way to work out the total?
Can you predict what would happen if I walked down the other side of the street instead (where all the houses have even numbers)?
Are there any patterns if I add together four house numbers instead of just three?
Or five house numbers?
Or...
Can you explain and justify the patterns you have noticed?
This task was used in a student webinar led by members of the NRICH team. You might like to watch the video footage of the webinar.
Getting Started
Student Solutions
This task was used in one of our student webinars, thank you everyone who took part. During the webinar, we received 199 submissions. Some are included in the published solution, but to see all of them together, click here.docx
Oak Class from Bishopsdown explained why their answers were odd:
We have found out that when you add three odd numbers together the answer will always be odd. If you add two odd numbers together, the answer will be even for example, 1 + 3 = 4. If you then add another odd number to the total the answer will be odd e.g. 4 + 5 = 9.
6AT from Castle Academy noticed that:
...all answers are in the 3 times table
We discovered that adding 3 even numbers have the same result.
Taking this idea further, Max from Holy Trinity Pewley Down sent in this observation:-
Zara, Rizni and Mia from St. Christophers made the following observations:-
and Alvic and Sophie also from St. Christopher's added the following:-
Josh from Bignold Primary School wrote:-
When you have any amount of consecutive numbers (such as 3, 5 and 7) when you divide your number by the amount of numbers you have you get a whole number, for example, 3+5+7=15 15÷3=5 and 5 is a whole number. Also, that number can be classed as the 'middle number' as 5 comes between 3 and 7.
Saying you did the conjecture with a different set of numbers, such as 3, 5, 7 and 9, if you add the up (3+5+7+9=24), and then divided it by the amount of numbers (24÷4=6) you would get 6, which is a whole number and can be classed as the 'middle number' as it is between 5 and 7 .
Ashlyn Roy and Grace from St Christopher's Bahrain produced a simple algebra for the 3 houses:-
Louis from Olga Primary noticed something similar adding 5 consecutive houses:
When you add 5 odd numbers together the middle number always is the [middle number] times by 5.
I think this is because with each time the two numbers before taking away 4 and the 2 after adding four cancelling out the 4 before so you end up with the answer.
Children at Thorpe Primary in Peterborough produced a display of their work on this task. It includes some pictures which help to explain why the sum of three consecutive numbers is three times the middle number, and why the sum of five consecutive numbers is five times the middle number:
Shriya from The International School Frankfurt in Germany explained this using algebra:-
If I add five house numbers, the result would be 5 times the middle number
because -2 would cancel +2 and -4 would cancel +4.
E.g. 2+4+6+8+10=30
The middle number is 6.
6×5=30
This is the same with odd numbers.
If I add four house numbers, the result would be 4 times the average.
E.g.1+3+5+7=16
The average is 4.
4×4=16
This is the same with even numbers.
Siddharth from Village School, Princeton Junction, NJ, U.S.A also looked at more than 3 numbers and developed a rule:-
I started by listing consecutive odd numbers and adding them up in a spreadsheet. I started with 3 numbers, then 4 numbers, 5 numbers and so on. I started to see a pattern. The middle number (when there were odd number of numbers) or middle numbers (when there were even number of odd numbers) were important because I could find a rule using them. Please see below. I found that the rule was
middle number X number of odd numbers [when there were odd number of odd consecutive numbers] (sum of the 2 middle numbers) X (the number of odd numbers/2) [when there were even number of consecutive odd numbers]
Using the rule, I determined the sum of first 19 odd numbers
(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37).
The middle number is 19, so 19 X 19 = 361. I then checked that my solution was correct by adding the 19 numbers using a calculator.
Using the rule, I determined the sum of first 20 odd numbers
(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39). The middle numbers are 19 and 21, so (19+21)X(20/2) = 400. I then checked that my solution was correct by adding the 20 numbers using a calculator.
I then started a spreadsheet for 3 consecutive even numbers, 4 numbers and 5 numbers and so on. I saw that the even numbers followed the same rules as the odd numbers. I think it is because every consecutive odd number or consecutive even number is 2 numbers apart.
Frida and Jake from Colneis Junior School thought about the sequences made by the totals. This is some of Frida's work:
Jake also tried adding numbers with gaps other than 2:
David from LAET also used the sequences formed by the totals, and used them to find formulae for the sums of the house numbers. David's work includes:
- exploration of the sums of different numbers of house numbers
- a formula for adding $n+1$ house numbers
- a proof of this formula for even numbers (which uses sums of arithmetic sequences)
- two proofs of this formula for odd numbers (one using induction, one using sums of arithmetic sequences
- deducing the formula for gaps other than 2 (continuing Jake's idea above)
Click to see each of David's proofs.
David - exploration of the sums of different numbers of house numbers:
Sum of 2:
$n_0$: 1 3 5 7 9 ...
sum: 4 8 12 16 20 24 ... $(2n_0+2)$
Sum of 3:
sums: 9 15 21 27 33 ... $(3n_0+6)$
Sum of 4: 16 24 32 40 ... $(4n_0+12)$
Sum of 5: 25 35 45 55 65 ... $(5n_0+20)$
$\sum_{r=k}^{k+3}{2r+1} = 4(k+3) = 4k + 12$ (as given by the 'formula' for four numbers)
David - a formula for adding $n+1$ house numbers:
$$\sum_{r = k}^{k+n}{2r+1} = (n+a_0)(n+1)$$ where $a_0=2k+1$ is the starting point
"The sum of the first $n+1$ house numbers (odd) from the house numbered $a_0$"
And for even numbers:
$$\sum_{r = k}^{k+n}{2r} =(n+a_0) (n+1)$$
David's proof of this formula for even numbers:
David's proof of this formula for odd numbers using proof by induction:
David's proof of the formula without using proof by induction:
\\
&=\frac{k+n+1}{2}(1+(k+n)\times2+1)-\frac{k}{2}(2+(k-1)\times2)\\
&=(k+n+1)(1+k+n) - k(k+1-1)\\
&=(k+n+1)^2 - k^2\\
&=(k+n+1-k)(k+n+1+k)\\
&=(n+1)(2k+n+1)\end{align}$$ where $2k+1 = a_0$
My formula concludes that:
$\sum_{r=k}^{k+n}{2r+1} = (n+1)(n+a_0)$ where $n$ is a positive integer,
and
$\sum_{r=k}^{k+n}{2r} = (n+1)(n+a_0)$ where $n$ is a positive integer.
Note: $\sum_{r=k}^{k+n}{2r+1} = \sum_{r=k}^{k+n}{2r} \not\Rightarrow 2r+1 = 2r$
David's formula for gaps other than 2:
Thank you to those who took part in the successful webinar and for the submissions that came in afterwards.
Teachers' Resources
Why do this problem?
This low threshold high ceiling problem is rooted in a real-life context where number patterns naturally occur, thus appealing to students' curiosity about the world. Students can go through the process of playful exploration, noticing patterns, making conjectures, and finding ways to justify their conjectures. The justifications can be numerical, visual or
even algebraic.
Possible approach
This problem featured in an NRICH webinar in June 2019.
Show the image of the house numbers.
"What do you notice? What mathematical questions might you ask?" Give students a short time to think on their own, before talking to a partner and then sharing with the whole class.
Here are some examples of ideas that might emerge:
The houses might be next door neighbours
They are all odd
The next one might be 13
The total is 27
"I added together some sets of 3 house numbers together as I walked past. Your challenge is to explore the different totals that I might have made."
Give students some time to try out some examples by adding together sets of 3 consecutive odd numbers, and then bring the class together and collect their totals on the board. In order to perceive patterns, it's useful to write out the three numbers used, as well as the total.
"What do you notice about all the totals we have found?" (You may wish to write their noticings on the board.)
They are all multiples of 3
They are all 3 times the middle number
Here are some possible representations that might emerge:
Using facts about odd and even numbers: When I add two odd numbers, the answer is even, so when I add three odd numbers the answer is even plus odd, which is odd. (This might be justified using a visual representation of odd and even numbers).
Numerical (using a generic example): 31 + 33 + 35, I take two off the 35 and add it to the 31 so I have three lots of 33. This will always work because the third number is two more and the first number is two less than the middle number.
Algebraic: I could write the three numbers as n, n+2 and n+4, so the total is 3n+6, which is 3(n+2). (Alternatively students could use n-2, n and n+2, giving a total of 3n.)
Visual:
If I have towers to represent each consecutive odd number, then each tower will be two cubes higher than the last one. I can move two pink cubes and put them on top of the blue tower to make three towers the same height.
Next, you might like to ask the class to come up with other mathematically interesting questions about consecutive house numbers. Here are some that might emerge:
What if I add together four house numbers instead of just three?
As before, invite students to explore one of their questions, making sure that they explain and justify any patterns they notice and conjectures they make.
Key questions
What is the relationship between two consecutive odd numbers?
How could you represent the three house numbers? Using multilink/squared paper/algebra?
Can you see anything in your examples that would work in exactly the same way if you used three different consecutive odd numbers?
What's the same and what's different when you add three consecutive even numbers rather than three consecutive odd numbers?
Possible support
Offer students some concrete representations of sets of three consecutive odd numbers, like the multilink representation above, or the equivalent picture on squared paper. Three Neighbours is structurally similar to this problem, but focuses on three consecutive numbers rather than three consecutive odd numbers.
Possible extension
Students could explore which numbers it is not possible to make by adding together neighbouring house numbers (and for a challenging extension, prove any conjectures they make).
You could also challenge students by giving them a particular number and inviting them to work out how many different ways it can be made by adding together neighbouring house numbers. For example, 600 can be made in eleven different ways!
Summing Consecutive Numbers has some interesting extension questions to explore, based on a similar task looking at consecutive numbers rather than consecutive even and odd numbers.