Sticky Dice

Throughout these challenges, the touching faces of any adjacent dice must have the same number. Can you find a way of making the total on the top come to each number from 11 to 18 inclusive?

Age

7 to 11

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

Throughout these challenges, the touching faces of any adjacent dice must have the same number.

In the picture above, three dice are joined with 4s between the blue and green, and therefore 3s between the green and red.

The total of the numbers on the top is 9 in each case.

The two arrangements will be thought of as the same even though they are in a different order and colour. This is true for all the challenges.

For the challenges below, you'll be using four dice.

Challenge 1

Find a way of making the total on the top come to each number from 11 to 18 inclusive, keeping to the same joining numbers throughout.

Here is an example:

Remember, any touching faces must be a pair of the same numbers.

Challenge 2

Using any joining numbers, find as many different ways as possible to make the tops add up to 12.

How can you be sure you've found all the possible ways?

Challenge 3

Arrange the dice in as many ways as possible, so the total of the tops of the left-hand pair is twice the total of the tops of the right-hand pair.

In the picture below, the top total of the left-hand pair is three times the right-hand pair, not two times so you cannot use this arrangement.

Challenge 4

Arrange the dice in as many ways as possible, so the pair on the left-hand side has a top product* which is double or treble the right-hand pair product.

In the picture above, the left-hand pair product is twelve times bigger than the right.

*The product of two numbers is one multiplied by the other. For example, the product of 3 and 5 is 15 because 3x5=15.

Here is a sheet of the activity.

Student Solutions

Alba from Barton C of E School in Cambridge, UK completed the first three challenges. Here are Alba's thoughts on each challenge:

First challenge:

The task was to make all the numbers from 11 to 18, I realised the smallest I could go to was 4 and the highest number I could go to was 24. I made 27 different ways, because there were lots of ways of making 13, 14, 15 and 20.

I spotted lots of patterns, I made sets of stairs.

My sticky numbers were 3, 4, 3.

That meant I only had dice options which were 1, 2, 5 and 6 because our sticky numbers are 3, 4, 3 so you can't use number 3 or 4.

If you have a big number and you want to make it one smaller then, for instance, you can swap a 2 for a 1 or 6 for a 5. Or if you wanted to get bigger you could swap a 1 for a 2, or a 5 for a 6.

I know 4 was the smallest because with only 4 numbers you can do 4 ones.

24 is the biggest because it's all of the highest number, 6, and you can only use 4 dice.

I made a table to make all the different ways you can make numbers from 4 to 24. (This table can be clicked on to make it bigger.)

You've successfully completed Challenge 1 by finding a way of making each number, but I think there are a few more ways of making some of these numbers if you would like to find all of the possibilities.

Second challenge:

This task was to find all the ways of making 12

Sticky numbers	Numbers I can use	How to make 12
1, 6, 1	5, 4, 3, 2	5, 3, 2, 2 4, 4, 2, 2 4, 3, 3, 2 3, 3, 3, 3
2, 5, 2	6, 4, 3, 1	6, 4, 1, 1 4, 4, 3, 1 3, 3, 3, 3
3, 4, 3	6, 5, 2, 1	6, 2, 2, 2 5, 5, 1, 1

There are 9 different ways to make 12. I know I have found all of the ways because I worked systematically. Systematically means I worked in order down the number line. (This table can be clicked on to make it bigger.)

Well done for working systematically, Alba! You're right, these are the nine different ways of making 12.

Third challenge:

This task was to find ways of making the two dice on the left-hand side add together to double the amount of the two dice on the right hand side added up.

Sticky numbers	Numbers I can use	Possible ways
1, 6, 1	5, 4, 3, 2	5, 4, 2, 3 (this is the only solution in this table that is incorrect - can you see what Alba might have meant to type?) 4, 4, 2, 2
2, 5, 2	6, 4, 3, 1	6, 6, 3, 3 6, 4, 4, 1 4, 4, 3, 1 3, 1, 1, 1
3, 4, 3	6, 5, 2, 1	6, 6, 5, 1 6, 2, 2, 2 5, 1, 2, 1 2, 2, 1, 1

The left-hand side has to add up to an even number so you can halve it.

This is very nearly a complete set of solutions - there are only two missing, both of which use the sticky numbers 1, 6, 1. I wonder what they could be?

Thank you, Alba, for sending in this solution. You have worked very hard on this!

Cassia, Edie, Nic, Kit from Atelier 21 Future School in the UK had some similar ideas for the third challenge:

We found out that there are 20 possible ways of completing Challenge 3. The first two dice have to add up to an even number because you can't do halves on a dice for the second two dice. The first two dice cannot add up to two because you cannot make one with two dice. When the first two dice add up to 8, it gives the greatest number of possibilities because it is in the middle.

Good ideas, and I agree that 8 is the most common total for the first two dice. I think some of your solutions use the same numbers in a different order, because I think there are only 11 different solutions for Challenge 3.

If anybody else would like to share their ideas about this task, please email us.

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