Copyright © University of Cambridge. All rights reserved.

We had well over 200 solutions sent in to this activity, the most we've ever had to read! Thank you all, there were lots that made interesting reading, particularly from those who described how they went about working on the challenge. The solutions also came from many parts of the world: the Czech Republic, United Arab Emirites, the United Kingdom, the United States of America and Australia. We are sorry we cannot include all the correct solutions but here are a few that give you a taste of how it was approached.

Year 6 pupils from St John Fisher Harrogate Magic Maths Club:

We started off by thinking of all the possible ways of making the totals. This took a long time.

We thought that it would be best to make the biggest totals first, using the bigger numbers to make them:

14 = 9 + 5, 13 = 6 + 7, 1 + 2 = 3, 4 + 3 = 7 and 8 + 0 = 8.

Some of us did it the other way round, making the smallest totals first, with the smallest numbers:

1 + 2 = 3, 4 + 3 = 7, 8 + 0 = 8, 7 + 6 = 13 and 9 + 5 = 14.

We could also come up with pairs randomly but it's quicker to use a strategy.

7 + 0 = 7, 5 + 3 = 8, 9 + 4 = 13, 6 + 8 = 14 and 1 + 2 = 3.

Ieuan from Riverside School, Prague in the Czech Republic wrote:

At first I was randomly picking numbers, and on my first attempt doing it I found a solution:

0 + 7=7 5 + 3=8 9 + 4 =13 6 + 8=14 2 + 1=3.

And then I tried using a system from then on, of adding a number to the smaller number then subtracting one from the smaller number, but it did not go very well because when I converted 5 + 3 to 4 + 4 I realised that you cannot do that.

Then I found out something quite clever that from one onward: each 2 numbers have the same amount of possibilities (I think by this Ieuan means that pairs of consecutive numbers have the same number of possibilities). For example 2 and 3 have two possibilities, 4 and 5 have have three, 6 and 7 have four, 8 and 9 have five and it goes on forever! So I wrote down all the possibilities for 7, 8, 13, 14, 3.

Then I shortened it, so if I use 13 as an example: 13+0, 12+1, 10+3, 9+4, 8+5, 7+6. Then I would take off 13+0, 12+1 and 10+3 and do that for all the rest! So when I had all the possibilities I did two attempts without succeding then I got one and I started explaining it on here, but I realised I had found the same as my first attempt.

Then I did one attempt and I found another: 3+0, 8+6, 9+4, 7+1 and 2+5.

Ellie in Year 5 at St. Nicolas CE Junior School, Newbury.

I found three solutions: For my first solution, I started with the highest envelope (14) and I thought it made sense to start with the highest card – 9. The number that went with 9 to make 14 was 5, so I had made one pair. Then I did 13, the second highest. As I had already used 9, I tried 8. To go with 8 to make 13, I needed 5 but that had already been used. So I had to use 7 and 6. Next, to make 7 from what was left, I had to use 3 and 4. Then 8 and 0 was perfect for 8 which left 1 and 2 for envelope 3 which worked. That was my first solution.

For my second solution, I tried a different solution for 14 which was 8 and 6. Then for 13 I had to use 9 and 4. There were two solutions to make 8 and this time I chose 5 and 3. Then 7 was made from 0 and 7 and 3 from 1 and 2 which all worked.

For my last solution, I used 8 and 6 for 14 again and 9 and 4 for 13. For 8, I used 7 and 1 which was different. To make 7, I used 5 and 2 (also different) and for 3, I used 0 and 3.

There were no other ways of making 14 so I knew I had found all the solutions.

So the numbers which could be inside the 8 envelope were 0 & 8, 5 & 3 or 7 & 1.

Rebekah from Thameside Primary School sent in a Word Document as follows:

From Angus of Canberra Grammar School Australia we had another interesting way of approaching this kind of problem

Again thank you, I hope that you learned something from reading these solutions.