# Centred Squares

One you may not have come across before is to do with squares. The sequence starts with one and then four points are put around it in a square:

Because the shape is growing bigger with new layers added around the outside and the centre staying the same, the sequence 1, 5, ... is called Centred Squares.

Here's the start of the third layer, but it is not yet completed:

Can you see how to complete the third layer?

You can now make your own 4th Centred Square.

Rather than saying that, for example, the second centred square has five points, instead we could say that the second Centred Square number is 5.

**Challenge 1**

**Challenge 2**

Explain how you did this.

**Challenge 3**

You can only use any Centred Square number once in any addition.

Explain how you chose the Centred Square numbers to use in your total.

**Challenge 4**

Explain how you did this.

**Challenge 5**

Remember you can only use any Centred Square number once in any addition.

**Challenge 6**

Thank you to everybody who sent in their solutions to this problem. Max, Alice, Catherine and Finn from Strand on the Green Junior School in the UK sent in this very clear solution to the first few challenges:

Challenge 1

We drew the first 5 centred squares to find them.

They are 1, 5, 13, 25 and 41.

Challenge 2

We noticed that to get the next centred square you add four more each time.

The 9th number would be 1 + 4 + 8 + 12 + 16 + 20 + 24 + 28 + 32

which is the same as 1 + (1 x 4) + (2 x 4) + (3 x 4) + (4 x 4) + (5 x 4) + (6 x 4) + (7 x 4) + (8 x 4)

If you add them together that makes 1 + (36 x 4) = 145

Challenge 3

We worked out the first 9 centred square numbers are 1, 5, 13, 25, 41, 61, 85, 113, 145

Max then carried on with the challenges.

He started with 145 because it is closest to the total.

211 - 145 = 66

You can make 66 using 61 + 5 or 41 + 25 so 211 is 145 + 61 + 5 or 145 + 41 + 25

Challenge 4

To get 212 you can just add 1 so 145 + 61 + 5 + 1 or 145 + 41 + 25 + 1

213 - 145 = 68 and we couldn't find a way to make 68 so tried starting with 113 instead.

213 - 113 = 100

100 - 61 = 39

39 - 25 = 14

1 + 13 + 25 + 61 + 113 = 213

Can you follow their reasoning to see how they worked out that last total?

We also received lots of solutions from the children at Ganit Kreeda in Vicharvatika, India. Well done to Sowjanya, Avyukt Ag., Adhrit, Achyuth, Ridhit, Niveditha, Pranit, Taran, Avyukt As., Rudransh, Abhiram, Nishika, Vidyut, Vishnu and Bhuvana for thinking so carefully about this!

Avyukt, Achyuth, Niveditha and Rudransh also noticed that to get the next Centred Square number we add four more each time, and they used this picture to work out why this happens:

In this picture, 'n' represents the number of new dots added along one side of the Centred Square number. For the second Centred Square number, that will be 2, and for the third Centred Square number, that will be 3, etc.

Niveditha used algebra to explain the pattern:

The number of dots added extra in the new figure is 2n + 2(n - 2) = 2n + 2n – 4 = 4n - 4 = 4(n - 1)

The other approach to count extra dots is: 4n – 4(corner dots) = 4(n - 1).

Niveditha is saying that each time, we are adding nearly 4 lots of n to the previous Centred Square number - but if we did that then we would count the corner dots twice, so we have to subtract 4. I think there might be another way of drawing this picture to show that we are always adding 4 lots of one less than n... Can you see a way of showing that in the picture above, the red dots added in the outer layer are four lots of three?

Vidyut, Sowjanya and Avyukt spotted a different pattern to help them solve Challenge 2:

Every Centred Square number except for 1 is sum of 2 consecutive square numbers. For example: 5=1+4, 13=4+9, 25=9+16

The children here are using 'n' and 'Cn' to represent general numbers, where 'n' can count which Centred Square number we are up to and 'Cn' can represent that Centred Square number. For example, with the third Centred Square number, n would be 3 and Cn would be 3^{2} + 2^{2}. Can you see how this formula can help us work out the ninth
Centred Square number?

The children used the following picture to think about why this pattern happens:

Using this picture, can you see why we add two square numbers together to get each Centred Square number?

The children worked collaboratively to solve Challenge 6:

Every Centred Square (CS) number is odd and ends in either 1 or 3 or 5.

A CS number always leaves a remainder 1 when divided by 4.

Or in other words, a CS no. is 1 more than a multiple of 4.

As 211 leaves 3 as remainder when divided by 4, we need to use either 3 or 7 CS numbers to get 211.

Secondly, as all CS numbers end in 1 or 3 or 5, the way we can get 1 as the units place digit in a sum of 3 numbers is:

5+5+1 and 3+3+5

Using this logic, we tried all possible combinations and found:

211 = 5+61+145

= 13+85+113

= 25+41+145

212 = 1+5+61+145

= 1+13+85+113

= 1+25+41+145

= 13+25+61+113

= 25+41+61+85

213 = 1+13+25+61+113

= 1+25+41+61+85

Thank you all for sharing your ideas with us!

Why do this problem?

This activity was designed for the 2015 Young Mathematicians' Award so it might be a particularly useful activity for a small group of your highest-attaining pupils to work on. It is a useful vehicle for developing systematic approaches. It can be used as an activity to encourage children to explain in written or spoken words what it is they have done.

Possible approach

Since this activity is aimed at the most confident mathematicians, there will not be much that you have to do. Just run (walk) through the introduction as set out and let the pupils declare their different ways of getting the first challenge completed.You may find it useful to give out copies of the challenge on this sheet.

Key questions

Tell me about how you are getting a solution for the challenge you are working on.(The main thing when encouraging the pupils to use their skills is to avoid saying things about what you notice and directing them in your way of attempting a solution.)