# Six Numbered Cubes

## Problem

**The aim of this challenge is to find the total of all the visible numbers on the cubes.**

We are using six cubes. Each cube has six faces of the same number.

The shape we make has to be only one cube thick. The shape on the left is built correctly, but the shape on the right would not be allowed as it is two bricks thick in places.

The total of the shape on the left is 70. Can you see why?

**CHALLENGE 1**

Start by making a staircase shape. An example is shown below:

a) What is the highest total you can make by using this staircase shape?

b) What is the lowest total you can make by using this staircase shape?

c) How did you calculate the totals for a) and b) above? Why did you choose the method(s) that you did?

d) Have a go at making a total of 75 using a staircase shape.

**CHALLENGE 2**

**any shape**of single cube thickness, what is the

**lowest total**you can make?

How can you be sure this is the lowest total whatever the shape?

Can the lowest total be found in more than one way? Justify your answer.

**CHALLENGE 3**

**any shape**of single cube thickness, what is the

**highest total**you can make?

How can you be sure this is the highest total whatever the shape?

Can the highest total be found in more than one way? Justify your answer.

**CHALLENGE 4**

If the cubes are arranged in a single vertical tower (like this)

then no matter what order the cubes are in, the total cannot be 80.

*This problem featured in a round of the Young Mathematicians' Award 2014.*

## Student Solutions

We had two lengthy files of explanation which are well worth looking at from Tom and Andrew who attend the British School of Paris. See them here: Andrew.doc or Andrew.pdf and Tom.doc or Tom.pdf.

Mali, who goes to St. Philip's School in Cambridge England, said:

The highest total is 83. The largest numbers 4, 5 and 6 must go on the very top to make the total as large as possible.

The lowest total is 64. Now, the largest numbers must go at the very bottom to have fewer faces visible.

To get 75 the ladder should be made of 4 on the top, 1 underneath, 2 at the bottom. Next column, the 3 at the top, the 5 at the bottom. Finally 6 on its own.

To find this solution, I cut pieces of paper with the number of visible sides on (5, 3, 3, 4, 4, 2), and pieces of paper with the numbers 1, 2, 3, 4, 5, and 6. I created pairs, multiplied them and add the total. I kept changing the pairs around until I found 75.

We had a lot of solutions sent in from Our Lady of Lourdes School: Stanley, Sean and Hedley; Ailbe; Chelsea; Gabriel, Fintan and Charlie; Catsaneman; Mia and Zoe; Mathias and Ethan, Comjen; Nansai; Ohene and Noeleen, Eene and Joe. They made paper cubes with the numbers on and experimented with different parts of the challenge - well done and thank you for your submissions.

Nathan from Cornelius Vermuyden in Essex, England sent in the following:

6 cubes stand vertically in a tower. Each cube has a number written all over it. For example one cube will have the number 1 on each of its six faces, and another will have 2 on every face and so on. So we have six cubes numbered 1 - 6.

These cubes are then stood in a vertical tower in any random order, like so....

A

B

C

D

E

F

Each letter can represent any cube as long as it is there only once. With a tower like this the cubes all have only four of their six faces showing except the top cube, which has five of its six faces showing. This means that we have every cube with four faces and an extra one at the very top of the tower.

So we can see that any six consecutive positive integers (not including 0) when multiplied by 4 (for the number of sides seen) and added together, equal above 80, the lowest being 84 with our current numbers 1-6. If we say A = 1, then adding the final face at the top of the tower will only increase the number made by 1 giving us our lowest possible total of 85. Therefore a total of 80 is impossible.

Algebraically this can be shown by representing the first of the consecutive numbers with n. Let's put this one at the top of the tower and have the other numbers go consecutively down the vertical tower.

n

n+1

n+2

n+3

n+4

n+5

We then need to allow for the faces on each of our cubes that are shown so we can write our expressions as”¦

5n

4(n+1)

4(n+2)

4(n+3)

4(n+4)

4(n+5)

We then expand the brackets and simplify this to”¦

5n

4n+4

4n+8

4n+12

4n+16

4n+20

Now we can collect like terms to give us 25n+60

Using 1 as our lowest positive integer, substitute n = 1 into the expression 25n + 60 to give 25 x 1 + 60 = 85

Once again, this shows that the lowest possible total is 85.

Sabine, Amelia, Sophie, Liam, Tom, and James, from Hardwick Middle School, sent in their ideas. Here is just a sample from the first three:

For question 1 we found that 78 was our highest total. We made this total by putting 6 at the top, 3 below it and 2 below 3. Next to 3 we put 5 and below that we put 1. Next to 1 we put 4.

For question 2 we got 64. We got this by putting 1 at the top, 5 below 1 and 4 below 5. Next to 5 we put 3 with six below. Next to 6 we put 2.

On question 1 we had to make six be shown the most, then five, then 4. we had to make 1 shown the least, then 2, then three.

For question 2 we had to make 1 show the most and 6 the least.

We had to do this otherwise the answer wouldn't be as low or as high as it could be.

For question 3 we had to get exactly 75 put of a staircase of cubes. It took us a while but we finally got the answer. At the top we put 2, below that was 6 and below that was 3. Next to 6 was five with 1 below it. Next to 1 was 4. That made 75.

This next solution was sent via the Wild.maths site which also has this challenge. It was from Ana G. in year 6 at Oaks Primary Academy and can be viewed as the pdf here.pdf .

Thank you all for your work, there was too much to show everyone's workings. I gather that it was generally enjoyed.

## Teachers' Resources

Why do this problem?

This problem gives pupils the opportunity to use knowledge and skills associated with spatial awareness, addition and multiplication, and to explain their thinking. It also involves keeping to rules that must be followed. The further they progress through the activity, the greater the opportunities for learners to use a whole variety of
problem-solving skills. The activity also opens out the possibility of pupils asking “I wonder what would happen if . . .?”

Pupils' curiosity may be easily aroused while trying to find solutions to the challenges.

### Possible approach

It would be good to demonstrate the kind of arrangements that are allowed as well as making those that break the rules for the pupils to decide on what is okay.

You may decide that you want the pupils to work in groups of three or four. One set of numbered cubes will be needed for each group. Having set them the first challenge, it may be sufficient to stop there. Challenges 2, 3 and 4 can be introduced straight away or left to another occasion.

It is worth noting that in the "steps" arrangement as shown on the problem page, the 5 and 6 both have four faces showing; the 1 and 4 both have three faces showing. So, the cubes that have the same number of faces showing can be swapped allowing for more arrangements to be possible.

### Key questions

How are you working out the totals?

How have you got to this arrangement?

Tell me about your shape.

How sure are you that ...?

### Possible support

Some pupils may require help with getting the cubes to stack. Those who are unable to record their arrangements could have them photographed.