Sealed solution

We had nearly 100 solutions sent in to this problem! Thank you to everyone who sent something in. Lots of these contributions were interesting to read, especially work from students who described how they went about working on the challenge. The solutions also came from many parts of the world, including Australia, China, England, India, New Zealand, Oman, Portugal, Qatar, Scotland, Singapore, Spain, UAE, USA and Vietnam. With so many good pieces of work we can't include them all, but here are a few to give you a taste of the approaches people took. Well done to you all!

Many of you sent in one correct solution, listing the pairs of cards in each envelope. Quite a lot of you considered whether there might be other solutions and even went on to find them. One approach to finding solutions was trial and improvement, as explained by Nilaa from Hymers College, UK:

I started with the 3 envelope and I just assumed it was 1 and 2 to start with, then for the 7 envelope I did 0 and 7 (just another educated guess!). I skipped the 8 envelope because it was subject to the question and then I did the 13 envelope, in there I put 4 and 9 (estimate) and for the last envelope, 14, I put 6 and 8 in because it added to 14. The remains were the cards 3 and 5 so I put those in 8.

I think the envelope could have the numbers 3 and 5 but there are definitely other combinations like 2 and 6, or maybe 1 and 7!

Like this one, we received several good explanations that said where guesses or assumptions had been made. This can be really helpful when we're reading a solution so we understand where a step came from. When you're working on a problem it can also be helpful as these are points that you might come back to to take a different approach.

Nilaa also illustrated their solution using colour to match up the cards with their total.  Among the submitted solutions we noticed several different ways of showing how the pairs of number cards matched up.

It's great that Nilaa recognised that there could be other solutions. Varna from Doha College in Qatar and Liam from New Hall School in UK found a different solution. Here is Liam's solution: 

To make 3, I used 0 and 3; for 7, 2 and 5; for 8, 1 and 7; for 13, 9 and 4; for 14, 6 and 8.

And Gauransh from the British School Muscat in Oman, Rio from Oeiras International School in Portugal and Elisia from St George British School in USA all found another solution:

1st envelope numbers 1, 2 ;

2nd envelope numbers 3 ,4;

3rd envelope numbers 8,0;

4th envelope numbers 6,7;

5th envelope numbers 9,5.

Like many of you, Elisia noticed how her choices for one envelope affected the numbers she could use for another envelope: 

For the first envelope the only options were 0, 3 or 1and 2, I realized that if I did 0 and 3 I would have less options for 2nd envelope like 6, 1 and 5, 2 , and then I would not have options for 3rd envelope number 8.

Some students used this idea to help them get started and narrow down the possibilities. For example, Sunny and Patrick from Dulwich College, Beijing, noticed that there were only two possible ways to make the total of 14, (5,9) and (6,8) so they started with this to limit their options. Maryam from Hollywood School in Australia also worked from the largest totals to the smallest. Other students like Zayd from Coombe Hill Junior School, England, started with the smallest total. Here's what Zayd wrote:

We did it in order from the smallest to largest numbers. We found all the possible bonds for each envelope number and we kept using the strategy and crossed off the numbers each time we used them. We started with the "3" envelope because it was the one with the least bonds. When we got to envelope 14 there was only two numbers left each time and they made 14 so we knew it was right.

Inside envelope 8 could be- 0 and 8, 3 and 5 or 1 and 7.

This made us wonder whether it's more helpful to start with the smallest numbered envelopes or to start with the largest numbered envelopes. 

Like Zayd and many other students, one approach to finding all solutions is to work systematically. Some students started this approach by listing all possibilities that make a total of 8, as shown here by Clark from the Maths Enrichment Club at Stanborough School in UK: 

Several students took the idea of working systematically further by listing the combinations that lead to the totals on each envelope. This was the first step by students from Ganit Kreeda, Vichar Vatika in India, who worked collaboratively on this problem. They built this table that listed all possible combinations of cards that give the totals on the envelopes.

The table shows that there are four options for the 8 envelope: 0 and 8, 1 and 7, 2 and 6, 3 and 5. However, as several students noticed, not every pair in the table can be part of a solution. As Maryam from Hollywood School in Australia commented:

While I was solving the sums, I realised some combinations of digits was not possible as I had used them earlier.

Varna from Doha College in Qatar also realised this and helpfully included an example where they had noticed that the pairings couldn't work:

1st trial set:

1. (0, 3)
2. (1, 6)
3. (2, 6)
4. (4, 9)
5. (5, 9)

However, this set uses 6 and 9 twice.

This means that once you have found all the ways of making each total, to find a solution you still need to work out how you can make all the totals at the same time. 

From their own table, Kanav from Sachdeva Public School in India noticed that 9 cannot be in the envelopes with totals 3, 7 or 8. They used this to organise the options into two cases to find the different solutions. Case 1 had cards 9 and 5 in the envelope with the total of 14 and and Case 2 had cards 8 and 6 in this envelope. Here is their argument for Case 1.

Case – 1 Taking cards 9&5 in the card which is having sum 14

Now in the envelope having sum as 13, the cards can only have numbers 6&7 as repetition cannot occur. 

Now, in the envelope having sum as 8, the cards can only have numbers 0&8 as repetition cannot occur. 

Now, in the envelope having sum as 7, the cards can only have numbers 3&4 as repetition cannot occur. 

Now, in the envelope having sum as 3, the cards can only have numbers 1&2 as repetition cannot occur.

Ha Anh and My Anh from Vinschool Metropolis in Vietnam also explained how they worked from their systematic lists to a solution, this time starting from the smallest total:  

We first listed the possible variations which will result in the sum in the envelope. After looking at all possible options, we then started looking at each one and comparing the other. For instance, if the first envelope were to have 1 and 2 inside, the second envelope can only have either 3 and 4 or 0 and 7, so on. We eventually got the the results that match and doesn’t overlap which are: 0 and 3; 5 and 2; 1 and 7; 9 and 4; 6 and 8.

Samuel and Afnan from the Maths Enrichment club at Stanborough School in UK and Freya, Sam and Morgan also from UK took a similar approach and found all solutions. Here are the solutions Freya, Sam and Morgan found:

Solution 1:
To make 8 use 0 and 8.
then
To make 3 use 1 and 2
To make 7 use 3 and 4
To make 14 use 9 and 5
To make 13 use 6 and 7

Solution 2:
To make 8 use 3 and 5
then
To make 3 use 1 and 2
To make 7 use 0 and 7
To make 14 use 6 and 8
To make 13 use 4 and 9

Solution 3:
To make 8 use 1 and 7
then
To make 3 use 0 and 3
To make 7 use 2 and 5
To make 14 use 8 and 6
To make 13 use 4 and 9

You might wonder whether there are any more solutions. Several students, including Freya, Sam and Morgan, not only found these three solutions, but explained why 2 and 6 (which is the only other way to make 8) can't be together in a solution. As Freya, Sam and Morgan said:

You cannot use 6 and 2 to make 8. You would need 0 and 3 to make 3, but then cannot make 7 out of the remaining numbers.

Well done to everyone who worked systematically through all of the different possibilities and also managed to explain why 2 and 6 doesn't give a solution. 

An interesting question came up in work that students at Ganit Kreeda, Vichar Vatika submitted. They asked "If you are allowed to open just one envelope other than 8, which one would you open?" They noticed that the numbers that make 7 in each of the three solutions are also different. Take a look at the solutions above. Is this the same for any other envelopes?

We hope these examples have given you some ideas about possible approaches to this problem. Well done again to everyone who submitted solutions to this problem.