In this, the children have to think about multiples and non-multiples of three. When a digit is placed on the grid, it must alternatively create a non-multiple of three and a multiple of three in the numbers that it belongs to horizontally and vertically. The same digit cannot be used twice. The rules are simple but a solution for a three-by-three square is hard to come by.

Here is an interactive version of the challenge. There are more detailed instructions below. On mobiles, you need to hold down on a digit and then it will allow you to drag it to a cell.

The first digit I add must create a number that is not a multiple of three. I'll use the digit 5.

5 | ||

The next digit added must create a number(s) that is a multiple of three. In this example, if I placed a digit in the top-middle box, it would create a two-digit number in the fifties that is a number of three. Therefore, I would have to use one of the digits 1, 4 or 7. I'll use 1. I now cannot use a 5 or a 1 again.

5 | 1 | |

Now, once again, I need to create a number that is not a multiple of three. I'm going to place a digit in the middle box. Therefore, it can't be a 2, 5 or 8 as that would create a multiple of three vertically. I'll go with 4.

5 | 1 | |

4 | ||

As we alternate, the next number needs to be a multiple of three. I cannot go in the middle-left box as that would create two numbers (horizontally and vertically) and with the digits already there, it is impossible for them to both be a multiple of three.

I will go in the bottom-left box and put in 3. It is not adjacent to any other digits so the number it creates is 3 as well.

5 | 1 | |

4 | ||

3 |

If I then make a few more moves, we reach the situation below:

5 | 1 | |

6 | 4 | 8 |

3 | 0 |

The next move is the eighth and I am required to create numbers that are multiple of three. I have the digits 2, 7 and 9 still to use. However, no matter which I try, the horizontal and vertical numbers that are created cannot be multiples of three. This is easily proved using the divisibility rule that the digital sum must be a multiple of three. Taking the middle-bottom box as an example, the digits along the bottom row add up to 3. Therefore, the middle digit would need to be 9. However, in the middle column, the digits add up to 5. That requires the digit 7 to make a multiple of 3. These two requirements are not compatible. This is the challenge!

The main focus of exploration will be trying to find solutions that work. It is a good idea to have a way to record the order that they are writing the digits in as nothing could be more frustrating than finding a solution and then forgetting the order.

There is a lot of scope for mathematical thinking. As shown in the example above, combinations chosen early on have an impact on what happens later. It is unlikely that a solution is achieved with a random selection of digits.

With the divisibility rule for three being that the digits are required to add to a multiple of three, the digits can be grouped in terms of the effect they have. For example, if we have the number 76, the digits add up to 13. As such, combining it with the digit 2 to make 15; 5, to make 18; or 8, to make 21 will have the same effect. This can be very useful knowledge for placing the last two digits when you are forced to create more than one number and have fewer options.

### Solution

The digits are placed in the following order:

1, 3, 7, 6, 2, 9, 4, 0 and 5

4 | 0 | 2 |

3 | 6 | 9 |

7 | 5 | 1 |

There are some interesting things to notice. The 3, 6 and 9 all being in the same row and all multiples of 3. Is that required? We are left with one column and two rows, for a total of three numbers, that are a multiple of three. Is that a requirement? Are there even other solutions?

Taking this further, there are a lot of possibilities.

**What if the first go required a multiple of three and the second go required numbers to not be a multiple of three?**

A simple switch round of the rule but, crucially, it makes the last move even more challenging as there are fewer combinations that result in a multiple of three than not. It may be that this is impossible.

**What if we used a different-sized grid?**

If we can successfully fill a 3x3 grid, would a 2x2 grid be easier, harder or impossible to create? Why? What about 4x4? Would each digit be allowed twice in that configuration? These are great ideas that they might ponder.

**What if we included diagonally-created numbers?**

This could either replace or be added to the existing rules and make things even more challenging.

**There are so many other ideas that could be explored.**

**Children's possible use of reasoning skills:**

## Search

When children initially pick the digits to use, there will be an element of Roaming but the fact that they always have rules that govern their placement means that Seeking is constantly required. This is an island that really promotes the idea of planning ahead. As such, there will need to be a lot of thought from the children in the numbers that they are creating.

Combing can be used by taking back moves and trying others. It is really helpful to explicitly identify this strategy to them so that they also use it in the future.

## Organise

Grouping the digits by the effect they have is an effective list that the children will find useful. That puts 0, 3, 6 and 9 together; 1, 4 and 7; and 2, 5 and 8.

As they are picking a digit to place, they can create lists of possible numbers that could be created. This will encourage the careful selection of numbers.

## Discover

After a few attempts, the children might declare that it is impossible as a conjecture.

With the solution above, I mentioned several things that I had noticed. These kinds of things form the likely conjectures that will come.

There are also the conjectures that can be formed in terms of reaching a certain position and then declaring it to be impossible to complete.

## Investigate

To a certain extent, the entire premise of the island is to investigate whether a solution is possible.

Selecting the placement and value of the first digit would warrant lots of investigation.

## Argue

There is a lot of scope here for children to explore why certain combinations on the grid make a donation impossible. A lot of mathematical thinking around the divisibility rule of three can be demonstrated in discussions of strategies.

## Explore

I've given a few examples above but I'm really scratching the surface of directions that the children might take things further from this starting point.

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