# Six Ten Total

## Problem

**Warm up:**check that the total of the faces showing is 84 when the dice are arranged as shown:

**To play the Six Ten Total challenge:**

- All the blue dice need to have the same number on the top.
- All the red dice need to have the same number on the top.
- There always needs to be six of one colour and ten of the other.
- There always needs to be a difference of two between the numbers on the blue and red dice.

**The main challenge**

What is the total for each of these arrangements?

What do you notice about your arrangements and the corresponding totals?

Explain what you notice. What else do you notice?

#### Taking it further

Can you prove any of the things you've noticed from the main challenge are always true?

Instead of having a difference of two between the numbers showing on each face of the blue and the red dice, choose a new difference.

Now, what totals can you find? What do you notice? Explain what you notice.

#### Final challenge

Can you predict - without having to make them - what numbers need to be on the faces that you can see on the blue and the red dice to make a total of 42?

Are there more ways to make a total of 42?

Explain your reasoning.

What happens if you change the number of blue and red dice yet keep the structure the same? It must be a square and both sets of dice must still make triangles.

Can you make 42?

Explain your reasoning.*This problem featured in a preliminary round of the Young Mathematicians' Award 2014.*

## Student Solutions

Ava, Pia and Sofia from Bute House School in the UK wrote:-

Our teacher asked us to notice and wonder about the patterns we could find in the Six Ten Total problem. We began by working out all the possible totals and then arranged them systematically. We noticed that all of the totals were multiples of 4. We also noticed that the totals increased by 16 as we increased each of the two dice numbers by 1.

We wondered if we could find a link with a different size of square, so we tested 2 x 2 squares with 1 of one value and 3 of another, following the same rules as before. We then explored 3 x 3 and 5 x 5... We predicted what the totals would be and how they would increase and then tested to see if we were right. Here are our results: doc or pdf.

We had the following from St Lawrence's Indigo Class in the UK:

Cian and Jacob

On one of our solutions we had a total of 60 from 5 in the 6's and 3 in the 10's. We found it okay to do. The biggest total you can get is 84. The lowest total is 24 using 1 and 3. The total of 60 took a while to find out. The total of 84 on the front of the problem sounded like not the biggest but it was. We got four solutions on our sheet of paper.

Keely and Alfie

We found out that there were six different ways of the solutions. We used a 6 to 1 dice to solve it. the totals that we found were 68, 60, 28, 36, 116 and 108. for 68 we used 10 fives and six threes also for 60 we used 10 threes and six fives, for 28 we did six threes and 10 ones, 36 was six ones and 10 threes, 116 was with six sixes and 10 eights, and for the last one 108 was 10 sixes and six eights.

Alex and Sianne

We found out our highest number was 84, and we made it using ten sixes and six fours then adding them together. Our lowest number was 28, and we made it by using ten ones and six threes. And these are some others we found out:

ten fives and six threes = 68

ten twos and six fours = 44.

We noticed that the difference between our two answers were all 8. Also the amounts were even.

Arnold, Freya and Joey from Eleanor Palmer Primary wrote the following - please look at their video.

We solved the Six Ten Total problem and took it further using Scratch! We

previously submitted a solution but then made a better video of our final explanation.

See our video here: http://vimeo.com/131883998

Max and Edward from Levendale Primary School, Yarm, Stockton on Tees UK wrote at length:

What are the possible arrangements when you choose your own numbers for the dice using the rules ? 1 and 3, 2 and 4, 3 and 5, 4 and 6.

What is the total for each of these arrangements?

For 1 and 3 - The total is 36 because (6x1) + (10x3) = 36

For 2 and 4 - The total is 52 because (6x2) + (10x4) = 52

For 3 and 5 - The total is 68 because (6x3) +(10x5) = 68

For 4 and 6 - The total is 84 because (6x4)+(10x6) = 84.

We noticed that every time the number on the face of each of the die increased by 1 the total increased by 16 because there are 16 dice and when they all change by 1 you add on 16 because that is the same as adding 16 to the previous total.

Can you prove any of the things you've noticed from the main challenge are always true?

Yes because there are 16 dice and each time the number on the faces of the dice increases by 1 the total still increases by 16 because you are adding 1 to each die. No matter what the difference of the dice it doesn't change how much the total increases.

Instead of having a difference of two between the numbers showing on each face of the blue and the red dice, choose a new difference. Our new difference is 3 and our two numbers are 1 and 4. The total is 46 because (6x1)+(10x4) = 46

For 2 and 5 - The total is 62 because (6x2)+(10x6)=62

For 3 and 6 - The total is 78 because (6x3)+(10x6)=78

Now, what totals can you find? What do you notice? Explain what you notice.

The totals are 46, 62 and 78. We noticed that the total still increased by 16 each time.

AND NOW

Can you predict - without having to make them - what numbers need to be on the faces that you can see on the blue and the red dice to make a total of 42?

Are there more ways to make a total of 42? Yes 2 and 3.

There are many other ways of getting 42. I wonder whether anyone else found some?

Explain your reasoning - This is correct because both numbers are factors of 42.

What happens if you change the number of blue and red dice yet keep the structure the same? It must be a square and both sets of dice must still make triangles.

Can you make 42? - No not with our 6x6 square.

Explain your reasoning. First we did this with 2 and 3 and our total was 92. And then we did it with 1 and 3 and our total was 56. Finally we did it with 3 and 4 and our total was 128. This proves that you cannot make 42 with our method.

This obviously caused a lot of interest even if you follow all the solutions above there are many more things to explore about this arrangement of dice. Have a go!

## Teachers' Resources

### Why do this problem?

This problem gives pupils the opportunity to use their skills of addition and multiplication in a problem-solving situation. It is also useful in expecting learners to persevere while exploring. It also requires pupils to explain what they notice and, when taking it further, to produce a proof.

### Possible approach

It would be good to have 16 dice for each pair/group to use to duplicate the given situation. Then ask them about getting the total and how they achieved that. (This can be very enlightening as to how the children, faced with an unusual calculation, go about finding a solution.) The rules need to be very clearly stated and checked with the pupils and then they can investigate other
arrangements.

Some may feel that they can quickly move on to recording their arrangements and so may not need the dice.

### Key questions

What have you noticed?

Tell me about this.

What will you do next?

### Possible extension

The final paragraph can form an extension for pupils who have confidently proved why their findings from the main challenge are true.

### Possible support

Some pupils may need help in keeping track of the different dice. A frequent reminder by asking "What were the rules?" may be necessary. Some pupils might also benefit from starting with a simpler version of the task using fewer dice.