A long strip of paper

Question

A strip of paper has $1024$ units in length and one unit wide, divided into 1,024 unit square. The strip is folded repeatedly. The first fold is done such that the right edge coincides with the left. The result is a strip $512 \times 1$ at double the thickness. Then, the right edge of this strip is folded such that the right edge coincides with the left, resulting in a strip $256 \times 1$ with four times the thickness. This process is repeated eight more times. After the last fold the strip became a pile of $1024$ unit squares. How many of these squares are below the square that was originally $942^{nd}$ square counting from the left?

Answer 1

After first fold the $942^{th}$ unit square will change its position, because $942>\frac{1024}{2} = 512$, so it'll be on the right half of the paper. Now the fold works like a mirror, so the distance from the half of the paper will be the same so the the $942^{th}$ unit square is $942-512=430$ unit square after the $512^{nd}$ unit square. So after the fold we need to the the following calculation $512 - (942-512) = 1024 - 942 = 82$, so after the fold the $942^{th}$ unit square will be on top of the $82^{nd}$ unit square

Now for the second fold we have: $82 < \frac{512}{2} = 256$, so its on the left side, so it won't change the position, but it'll be on second level and will have two more unit squares on top of it.

For the third fold we have $82 < \frac{256}{2} = 128$, again our square unit won't change it's position, but now it'll have 6 unite squares on top of it and 1 under it.

For the fourth fold we have $82 > \frac{128}{2} = 64$. So this time it'll change its position and now it'll be on the $128-82=46$, which means it'll be on the $46^{th}$ position. But now take a look at the level it is. Note that after the fold the 6 unite square that were on top of it, now are under it. Add them to the 8 of the left part of the length, and we have that there are 14 squares under it and 1 on top of it (this is the one that was originally under it).

For the fifth fold we have $46 > \frac{64}{2} = 32$. So the position this time will be $64-46=18$. Now for the level we have to do the fliping. Now there will be 17 unit squares under it (16 of the left side, and 1 that was on top of it) and 14 on top of it.

For the sixth fold we have $18 > \frac{32}{2} = 16$. So again it'll change position and its new position will be $32-18=14$. And for the level we have 46 unit squares under it (32 of the left part and 14 that were on top of it) and 17 on top of it.

For the seventh fold we have $14 > \frac{16}{2} = 8$. It means it'll change position and its new position will be $16-14 = 2$. While its level will be 81 squares under it (64 from the left side and 17 that were on top of it) and 46 that are on top of it.

For the eighth fold we have $2 < \frac{8}{2} = 4$. It means that it won't change its position, but we after the fold there will be new 128 squares on top of it. So there will be 81 squares under it and 174 on top of it.

For the ninth fold we have $2 = \frac{4}{2} = 2$. But because the fold happens after that square, again our square will be on the left side. So its position won't be changed, but its level will. After th fold there will be 81 square under it and 430 on top of it.

And for the last fold it's on the right side so it'll change it's position to 1, while there will be $430+512 = 942$ unit square under it and $81$ on top of it.

So this means that our square will be on the $943^{th}$ position, counting from the bottom after the ten folds.

I must admit such a long journey for a slight movement. Anyway, maybe this brute-force method won't satisfy you, but I don't think there is any alogrithm that can help you.to find it faster.