This island shows that you do not need large numbers to make something challenging. All of the additions here are three-digit ones yet children will have to really consider how they use regrouping across columns to minimise the number of steps used in their sums.

Here is a 3-digit addition:

465 + 372 = 837

We have managed to created an answer that has the digits 8, 3 and 7. Our aim is to swap digits in our addends until we have a set of sums where each digit appears in at least one of the answers.

For example, I could swap the six used in the tens column of 465 for an eight to create a new addition:

485 + 372 = 857

I have now managed to create a sum that contains the digit 5 (as well as 8 and 7). Across my two additions, the digits 3, 5, 7 and 8 can be found in my two answers. I still have to find a way for my answers to include the digits 0, 1, 2, 4, 6 and 9.

How many times do I need to swap digits to achieve that?

**What if I started with a different addition?**

Is there a strategy that I can develop no matter what addition I start with? Is there a maximum number of swaps that would need to be made no matter the starting numbers?

**What if we could not swap a digit into the addends that had already been used?**

This is the main challenge that I want the class to work towards. By gradually adding the constraints and giving them a chance to explore first, I am trying to ensure that it is accessible. Otherwise, this might seem impossible when first attempted.

With this problem, if I start with an addition, for example the one I started with:

485 + 372 = 857

I have now used the digits 4, 8, 5, 3, 7 and 2 in my addends. I cannot swap those digits in. I can only swap in 1, 6, 9 and 0 as they have not been used. That means I have five additions in total to try and create a set of sums (answers) that contain all 10 digits. It is possible!

New Digits Used | Digits Available | Addition | New Digits in Sum | Digits Left to Make |
---|---|---|---|---|

2 3 4 5 7 8 | 1 6 9 0 | 485 + 372 = 857 | 5 7 8 | 1 2 3 4 6 9 0 |

1 | 6 9 0 | 481 + 372 = 853 | 3 | 1 2 4 6 9 0 |

6 | 9 0 | 481 + 672 = 1053 | 10 | 2 4 6 9 |

One of the trickiest aspects of this is choosing a starting calculation that is possible. In finding a couple of possibilities, I had to really think about the order of the digits I wanted to construct from the sums. As such, it might be that you provide this. There will be others but click below to see one that can work if you would like.

## One starting calculation that can lead to a solution

124 + 803

Using this starting point, there is a solution that is possible if you would like to see it. Using that solution, I was able to also come up with other, similar, solutions.

## Three similar solutions and the strategies involved in constructing them

124 + 803 = 927 |

124 + 863 = 987 |

524 + 863 = 1387 |

594 + 863 = 1457 |

597 + 863 = 1460 |

467 + 820 = 1287 |

167 + 820 = 987 |

567 + 820 = 1387 |

567 + 890 = 1457 |

567 + 893 = 1460 |

167 + 820 = 987 |

467 + 820 = 1287 |

467 + 890 = 1357 |

467 + 893 = 1360 |

567 + 893 = 1460 |

With the first two, four of the resultant sums are the same and the first addition is to get the leftover digit – two. The third is very similar. Instead of making three and then using regrouping with the next column to bump it up to a four, I made two and then bumped it up to a three.

In all of them, I used the sum of 987 and then altered a column at a time to make the remaining digits. Is this the only way of doing it? I am not sure. With three of the additions requiring us to make two new digits, regrouping is key. I started with the principle of having no columns regroup and then altered each column one at a time to make them regroup. Starting with high-value digits like with 987, seems essential to achieve this. I also left pairs of digits to make in each column. A three and a four in the hundreds (or a two and a three), and a five and a six in the tens. I then looked at how I could find a way to make the remaining digit.

An interesting thing to explore might be why having 789 is easier than 987. I am not sure whether using 987 is possible even though it feels like it should be. I am conjecturing that it is not.

I wonder how many solutions there are. Any solution can be reversed so that is two. I’d love to hear if you find any more.

**What if we used subtraction?**

This is an inevitable new idea to investigate following the use of addition. It is even more problematic as we do not have the thousands column available. But it is ok if it is not possible. It could still be investigated and explained.

**What if we used three addends?**

Instead of two three-digit addends, we could have three two-digit addends. This does only give us three columns to work with in the sum and so constructing all ten digits in only five calculations is possibly impossible but it could be something to explore.

**What if we could swap the digits adjacent to each other in the addends?**

This is an interesting idea. If we only had two three-digit numbers and were not allowed to swap for any new digits, could all ten digits be created in the answers? In how many steps? It draws in some nice thinking in terms of ensuring we have a balance of even and odd and high and low digits to use. I think I might turn this one into a new post as I think there is a lot investigation possible from it.

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

## Search

Whilst there are limited options, the big choice is in the initial choice of digits and their position. It might be that they have to Roam initially to understand how it will be possible to create a solution using only five additions.

However, Roaming alone is unlikely to lead to a solution. This is why reflecting on what they have created is as important as the creating of additions itself.

## Organise

It is very useful to have a list of digits left to use and a list of digits that have been created.

However, this is an island that will likely require many attempts before a solution is found. A table, like the one above, could be useful as long as it does not discourage multiple attempts.

## Discover

The key here is understanding the necessary conditions for the starting numbers to make it possible. There is a lot of thinking required and the clever use of regrouping across columns is necessary to build a solution.

## Investigate

This is an island that really promotes the careful thinking of strategies and then investigating whether they are possible. After an initial period of Roaming, if the children discuss why they might be struggling to create solutions, it will hopefully lead to strategies that they can use and investigate.

## Argue

As they are unlikely to come across a solution randomly, they can use their Arguing skills to explain why most combinations do not work. What is the difficulty being presented here and what can they do about it? Encourage discussion of how many new digits can be changed in the answer with each swap of digits. This will lead to strategies that could work.

## Explore

This island really utilised the Explore skills to add depth and thinking to the problem. It shows how we can gradually build in challenging constraints to build in a lot of mathematical thinking.

There are also lots of possibilities that we can explore from this starting point.

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