I solved these problems by starting with the basics: firstly, I realised that to make each question easier, you need to start with the larger numbers to make sure you have enough numbers to go in between them. Over the time I was solving these problems, I used quite a lot of mathematical trial and error.
For the first question: I began with the number 2, I put both the other numbers in between the number 2’s, therefore I knew that there would be two numbers between them and I could put the other pair of each number either side, so there would be one number between each 2. The answer is now 121323. As you can see, there is 2 numbers between the 2’s and one number between the 3’s and the 1’s. If I wrote this in algebraic terms, depending on odd and even numbers, it would be xyxxyx. The x=odd and y=even.
When I did the next question I used what I already knew. I started with the biggest number’s [3] and I used the numbers 121 and put them in between, this made what seemed like complicated things easy, by now I knew that there would be three numbers separating the 3’s and one number separating the 1’s. However I was still missing the other 2 in the number sandwich so just by putting it adjacent to the number 3, the problem was solved! My answer to this problem was 312132. This number matched the criteria for the question. Converting it to algebra again, it would be xxyxxy.
Following on to the third question,I used the two answers I had already got. Using these answers gave me a rough idea of what the answer should be like, because I was just increasing the answer by two of the next consecutive number. Although this might sound easy, it requires quite a bit of mathematical knowledge. I began with the 4’s, I placed 131 in between, therefore, there was one number between the 1’s. After that, I added a two to the right of the 4. Now, in my number sandwich so far was four numbers between the 4’s. I only had one three and one two left to place in my sequence. I knew that if I put the three to the left of the first four, there would be three numbers separating the 3’s, however now, I just placed the last four at the start of the number sequence. This formed my full sequence, applying to all the expectations/rules of this question. My answer was 23421314, which also could be written as yxyyxxxy.
Finally, for the most complex question, throughout the time I was constructing my number sequence, I used quite a lot of trial and error! I used a grid with 14 spaces on it to help me solve this number sandwich out. I began with the 7’s, putting one at the start and then counted on 8 spaces, because this meant that there would be seven spaces between them. Next, I half guessed and put for the second, third and fourth numbers 131. Following that, I started to work with the second largest number: 6, I positioned the first of the pair on the next spare space [the fifth] and then situated the second 6 seven squares on. After that, I decided to position the 4’s in place. I chose to place this one on the sixth empty spot, following my previous process, I moved on 5 places. I had five slots left and I had these digits left: 3, 5, 5, 2, 2. My next move was calculating which spaces the 5’s could go in, I managed to do this by finding out which two squares had five spaces in between. This worked, so that is where I put them. I then put the 3 four spaces away from the first three, and finally, I positioned the last two numbers [2’s] in the last two gaps to complete my number sandwich! My complete answer was this: 71316435724625, which in algebraic terms would be xxxxyyxxxyyyyx.
I tried out which numbers I could use to make a complete sandwich:
I could not do it with just 1
I could not with 1 and 2
I could with 1,2 and 3
I could with 1,2,3 and 4
I could not do it with 1,2,3,4 and 5
I could not do it with 1,2,3,4,5 and 6,
But I could do it with1,2,3,4,5,6 and 7!