Folding ?

[Hypatia001]

Once again i am stumped on a puzzle asked by my 10 year old. I have physically created the design and get an answer of 3... can anyone help me to see if this is correct?

The reason i ask is because when i try and do it mathematically i came out with a different answer !!!

 

[lev888]

Once again i am stumped on a puzzle asked by my 10 year old. I have physically created solution the design and get an answer of 3... can anyone help me to see if this is correct?

The reason i ask is because when i try and do it mathematically i came out with a different answer !!!

Please post your solutions.

 

[Zermelo]

So if you chose to keep the three fixed in top, there are two ways you can continue: first fold the right half (4 and 5) and then the left half (1 and 2) or the other way around. Note that you cant do 3 4 1, for example, you first have to do the half that you started with. So, you can do 3 4 5 1 2, 3 4 5 2 1, 3 5 4 1 2, 3 5 4 2 1. You figure out the rest ( if you start with the left half). I hope that I didn’t miss any rules about folding. Mathematically, you can use permutations without repetition

 

[Hypatia001]

Thanks.

If only the 3 can stay on top.... and i have actually made a little paper version of this then the only solutions i get are 32145, 34512 and 34521.

Whqt am i missing ?

 

[Hypatia001]

Sorry.... 32145, 34521 and 32154

 

[Hypatia001]

So if you chose to keep the three fixed in top, there are two ways you can continue: first fold the right half (4 and 5) and then the left half (1 and 2) or the other way around. Note that you cant do 3 4 1, for example, you first have to do the half that you started with. So, you can do 3 4 5 1 2, 3 4 5 2 1, 3 5 4 1 2, 3 5 4 2 1. You figure out the rest ( if you start with the left half). I hope that I didn’t miss any rules about folding. Mathematically, you can use permutations without repetition

So here is an update... we found 8 ways....

32145
34521
32154
34512
31245
31254
35421
35412

Are there any more you guys can think of ?

 

[Dr.Peterson]

So here is an update... we found 8 ways....

32145
34521
32154
34512
31245
31254
35421
35412

Are there any more you guys can think of ?

That should be it. I would have ordered them in a way that would make it easier to see that you haven't missed or duplicated any, but it is clear that the number will be 2^3 = 8.

You can have either 12 or 21, and either 45 or 54, and these can be in either order relative to one another. Each pair must be together. So we have 2*2*2 choices. This is essentially what Zermelo said.

 

[Hypatia001]

That should be it. I would have ordered them in a way that would make it easier to see that you haven't missed or duplicated any, but it is clear that the number will be 2^3 = 8.

You can have either 12 or 21, and either 45 or 54, and these can be in either order relative to one another. Each pair must be together. So we have 2*2*2 choices. This is essentially what Zermelo said.

Thanks. Order was never my strong point .... although am trying to teach my daughter to keep things neat and tidy.

 

[JayJay]

Here's two more. Fold 2 under 3, then 1 under 2. Fold 4 under 1, then tuck 5 between 2 and 1. This gives 3,2,5,1,4. Exchange left / right to give 3,4,1,5,2.

 

[Hypatia001]

Here's two more. Fold 2 under 3, then 1 under 2. Fold 4 under 1, then tuck 5 between 2 and 1. This gives 3,2,5,1,4. Exchange left / right to give 3,4,1,5,2.

Yes you are right !! @Dr.Peterson is there a mathematical way to solve this ?

 

[Dr.Peterson]

Here's two more. Fold 2 under 3, then 1 under 2. Fold 4 under 1, then tuck 5 between 2 and 1. This gives 3,2,5,1,4. Exchange left / right to give 3,4,1,5,2.

Yes, I missed the tuck, by doing too much in my head.

You could also think of this as folding 2-1 behind, then folding 5 behind 1, and then folding the whole assemblage.

Yes you are right !! @Dr.Peterson is there a mathematical way to solve this ?

No, other than being systematic and open to unexpected possibilities, if you consider that to be math. (I do! Math is not just formulas.)

 

[Hypatia001]

Yes, I missed the tuck, by doing too much in my head.

You could also think of this as folding 2-1 behind, then folding 5 behind 1, and then folding the whole assemblage.

No, other than being systematic and open to unexpected possibilities, if you consider that to be math. (I do! Math is not just formulas.)

That makes 10 !! Thanks agina for the help @Dr.Peterson @JayJay @Zermelo