Problem Teachers' Notes Hint Solution

Three Dice

21 is multiple of 3, therefore if one sum is multiple of 3, other sum must be too.
21 not multiple of 4, therefore both sums can't be multiples of 4.
28 not multiple of 3 therefore not possible.

Previous solutions up to Sept 2012 were:
We had some extremely well-explained solutions to this problem - thank you for sending them in. Red Kites Class from Holton St Peter Primary said:
We really enjoyed solving this problem.
We noticed that the top number and the bottom number of any dice always added up to $7$. For example, $2$ and $5$, or $4$ and $3$.
Then we looked at the sum of the top numbers and the sum of the bottom numbers. We noticed that they always added up to $21$. For example, $6+5+3=14$ on the top and $4+1+2=7$ on the bottom. $14+7=21$.
We did lots of examples to make sure that we were certain.

Well done! I like the way you have found lots of examples that fit your rule. Mehwish, Becky and Amber from Oakwood also tried lots of examples and found that they all made a total of $21$. When you discover a rule like that, the next stage might be to ask 'I wonder why that happens?'. Ryan from St Ebbe's Primary has started to explain for us:
If you keep throwing dice, you will eventually find that the sum of the top sum and the bottom sum adds up to $21$. This is because the top and bottom number of a die is always $7$. And since there are three dice, $7\times3=21$.
With four dice it is $7\times4=28$.
With three dice the top and bottom sums can be a multiple of three but not four because three goes into $21$ ($21\div3=7$) but four doesn't ($21\div4=5.25$).
And following on with four dice, $28\div4=7$ (that works) and $28\div3=9.3333.......$ (that doesn't work).
So it pretty much depends on how many dice you have.

Well done, Ryan, some clear reasoning here. Bryce, Cooper, Silas, Taylor, Aaron, Sydney, DaeLen, Dylan, Brandon and Byron from AIG also looked at having more than three dice:
Brandon and Aaron realised that the number of dice affected the sum. So, as a group it was decided that the number of dice we had times seven would be the sum and it would be a multiple of $7$.
We tested this theory using three, four and five dice. We then double checked it with one and two dice. We found that our ideas were correct.

Krystof also sent in a good solution - well done.
Stuart, Rebecca, Rhys and Eva who go to Myland Primary School clearly worked very hard on this problem too. They sent in a very full solution and looked at dice with different numbers of faces:
When we throw three dice and add up the top numbers and the bottom numbers, one sum is even and the other is odd. The pairs of totals are:

$3$ & $18$
$4$ & $17$
$5$ & $16$
$6$ & $15$
$7$ & $14$
$8$ & $13$
$9$ & $12$
$10$ & $11$

We noticed that the top and bottom totals all add up to $21$. Opposite sides of the dice add up to $7$. So when we add the numbers on opposite sides of three dice we get $7+7+7 = 21$.
If the top numbers on three dice add up to a multiple of $3$ then the bottom numbers also add up to a multiple of $3$ because $21$ is a multiple of $3$.
Totals on four ordinary dice $= 7 + 7 + 7 + 7 = 28$
Totals on five ordinary dice $= 28 + 7 = 35$

On dice the highest number is always opposite the lowest number, the next highest is opposite the next lowest. So on an eight-sided dice numbered $1$ to $8$ the opposite sides would add up to 9 and the total on three dice would be $27$.
On a 100-sided dice numbered $1$ to $100$ the opposite sides would add up to $101$ and the total on three dice would be $303$.
We investigated our ten-sided dice, numbered $0$ to $90$:
Example 1:
$90 + 80 + 70 = 240$
$0+ 10 + 20 = 30$
Total = $270$

Example 2:
$40 + 50 + 60 = 150$
$50 + 40 + 30 = 120$
Total = $270$

The opposite sides of the dice add up to $90$ so the total is always $90+90+90 = 270$.

What fantastic work - I love the way you have begun to ask your own questions and explored things that interested you. You also worked very sytematically to find all the possible combinations of totals on three dice, which is very helpful in answering this problem. It makes me wonder ... will a dice always have an opposite face, no matter how many faces it has?
Well done everyone!