The question is simple: How many times do we have to fold a sheet of paper so that, by incresing its thickness, we cover the distance from Earth to the Moon?

Let’s all assume that the thickness of the sheet of paper is 0.1 mm and the distance Earth-Moon is 384.000 km.

Hex| PUZZLE MASTER | Profile February 3rd, 2010 - 3:32 pmAmazingly enough, 42 folds would exceed this distance.

0.1 x 2^n >= 384,000,000,000

n >= ln(3,840,000,000,000)/ln(2) ~= 41.8

cheddarmelt| Profile February 3rd, 2010 - 7:28 pmMore times than the paper will tolerate–it’s an impossibility. The sheet of paper would have to be longer than the Earth-Moon distance just for the outermost layer to wrap around the rest of the folded wad.

michaelc| Profile February 3rd, 2010 - 8:21 pmA short and sweet one. I think this was on Mythbusters perhaps awhile back. You can’t physically fold a sheet of paper more than about 9 or 10 times. Take a sheet of notebook paper and see how many times you can fold it in half.

For the fun of thought however, let’s say we fold this sheet in in half with our minds! 42 Times will get you 439,804 Km. 41 times leaves you with only half of that.

For fun’s sake, let’s say the final fold left a face area of about 1mm x 1mm. So that’s a volume of 1mm x 1mm x 439,804km. Which is about 4.4 x 10^11 cubic mm. This implies you need a piece of paper about 0.1mm x 2,097 m x 2097 m to have enough paper to make the fold!

Hendy| Profile February 3rd, 2010 - 9:30 pmDistance in mm is 384000 x 1000 x 1000

384,000,000,000 mm

00 384000000000.0

01 000000000000.1 x 2 = 000000000000.2

02 000000000000.2 x 2 = 000000000000.4

03 000000000000.4 x 2 = 000000000000.8

04 000000000000.8 x 2 = 000000000001.6

05 000000000001.6 x 2 = 000000000003.2

06 000000000003.2 x 2 = 000000000006.4

07 000000000006.4 x 2 = 000000000012.8

08 000000000012.8 x 2 = 000000000025.6

09 000000000025.6 x 2 = 000000000051.2

10 000000000051.2 x 2 = 000000000102.4

11 000000000102.4 x 2 = 000000000204.8

12 000000000204.8 x 2 = 000000000409.6

13 000000000409.6 x 2 = 000000000819.2

14 000000000819.2 x 2 = 000000001638.4

15 000000001638.4 x 2 = 000000003276.8

16 000000003276.8 x 2 = 000000006553.6

17 000000006553.6 x 2 = 000000013107.2

18 000000013107.2 x 2 = 000000026214.4

19 000000026214.4 x 2 = 000000052428.8

20 000000052428.8 x 2 = 000000104857.6

21 000000104857.6 x 2 = 000000209715.2

22 000000209715.2 x 2 = 000000419430.4

23 000000419430.4 x 2 = 000000838860.8

24 000000838860.8 x 2 = 000001677721.6

25 000001677721.6 x 2 = 000003355443.2

26 000003355443.2 x 2 = 000006710886.4

27 000006710886.4 x 2 = 000013421772.8

28 000013421772.8 x 2 = 000026843545.6

29 000026843545.6 x 2 = 000053687091.2

30 000053687091.2 x 2 = 000107374182.4

31 000107374182.4 x 2 = 000214748364.8

32 000214748364.8 x 2 = 000429496729.6

33 000429496729.6 x 2 = 000858993459.2

34 000858993459.2 x 2 = 001717986918.4

35 001717986918.4 x 2 = 003435973836.8

36 003435973836.8 x 2 = 006871947673.6

37 006871947673.6 x 2 = 013743895347.2

38 013743895347.2 x 2 = 027487790694.4

39 027487790694.4 x 2 = 054975581388.8

40 054975581388.8 x 2 = 109951162777.6

41 109951162777.6 x 2 = 219902325555.2

42 219902325555.2 x 2 = 439804651110.4

On the 42nd fold, the paper thickness would exceed the distance to the moon.

——————————————————

Here’s a question back at the group.

Assuming a machine could be made large enough to fold this piece of paper (and that paper folded will not bunch up so badly that it could not be folded no matter the abilities of the machine), estimate how large the paper would have to be (km^2) to have a width and depth of about 1 square kilometer?

BTW, I don’t have the answer, but it would be interesting to see how everyone arrives at their answer.

Blusummers13| Profile February 3rd, 2010 - 10:41 pmEither I did something wrong, or it’s a surprisingly small number. I get 41.66 or 42 folds since you can’t have a part of a fold.

My equation was 0.1*2^x = 348,000,000,000mm

Even though I know that the equation is exponential, 42 folds just still seems too small to me.

kasabubu| Profile February 3rd, 2010 - 11:44 pmSo far no one has fold a piece of paper more than 12 times. So maybe we should stack it instead ;-)

Here is an interesting link regarding the folding problem:

http://www.damninteresting.com/never-say-never

munna| Profile February 4th, 2010 - 2:36 amEach time we fold the paper, its thickness is doubled. This can be viewed as an Geometric progression [i.e. a, ar, a(r^2), a(r^3), … a(r^n)] where

^ denotes: to the power of

a = thickness of the paper = 0.1mm = 10^(-7) km

r = 2 (because thickness is doubled on each fold)

n = the number of folds

Let say at nth fold, the thickness of paper is equal to the distance between Earth and Moon i.e. (384 * 10^3) km. Hence we get:

a(r^n) = 10^(-7) * 2^n = 384 * 10^3

=> 2^n = 384 * 10^10

Apply log (base 10) and we get: n = (log384 + 10) / log2 = 41.8042433884

Since number of folds is a whole number, we will round the value to 42 (which is the answer).

i,e, we need to fold the paper at least 42 times so that its thickness covers the distance between earth and moon.

the_god_dellusion| Profile February 4th, 2010 - 6:30 ami think there are 2 answers…

1. You can only fold a piece of paper 7 times(i think or somewhere round there) then it can no longer be folded. this applied to any size paper.

2. make equation: 3840000000=1*2^x – where x is number of times folded. taking logs on each side and simplyfying you get;

x=log(3840000000)/log (2)

i dont have a calculator….

aaronlau| Profile February 4th, 2010 - 8:44 am42 folds?

Though i remembered some where mentioning that paper can only be folded a maximum of 8 times.

dhunnapotha| Profile February 4th, 2010 - 9:21 amlog(384 * 10^7) to the base 2 ?

engjs1960| Profile February 4th, 2010 - 10:18 am42 times will do it.

Shawn| PUZZLE GRANDMASTER | Profile February 4th, 2010 - 11:01 am43 folds

Assuming perfect folds with no air gaps between the sheets as the folds get thicker

thickness of paper = 0.1mm = 0.0000001km

distance earth-moon = 384,000km

Fold #1 = 0.0000002km

Fold #2 = 0.0000004km

Fold #41 = 219,902km

Fold #42 = 439,805km

So how big would the initial sheet of paper need to be to get 42 folds? I conducted an experiment with varying paper sizes:

5.5 x 8.5″ = 7 folds

11 x 17″ = 8 folds

22 x 34″ = 9 folds

If this pattern of (1) additional fold for each doubling in paper size holds true for larger sizes of paper, then in order to fold a piece of paper 42 times, the starting dimensions would have to be:

4,800,074km x 7,418,297km

Since the circumference of the earth is ~40,000km, good luck laying out that sheet of paper for the first fold – we will need an orbiting woodpulp mill in space, finally something useful for the int’l space station to do!!

ALSO:

Estimates for the number of sheets of 8.5″x11″ paper that can be produced from one tree vary. I found estimates ranging from 6,000 sheets to 80,000 sheets; I used 10,000 sheets per “standard tree.” That means that our one gigantic sheet would require the destruction of about 21.5 BILLION TREES.

Total # trees of all kinds estimated on earth as of 2005: 400 billion. WE CAN DO THIS, PEOPLE!!!!

Mashplum| PUZZLE MASTER | Profile February 4th, 2010 - 11:21 amI would have to be an accordion fold, but even then, as the width of each pleat approaches zero and the number of folds approaches infinity, the total thickness approaches the original length of the paper. So unless the unfolded paper already reaches the moon, this isn’t happening.

james turner| Profile February 4th, 2010 - 2:25 pmLet’s define some variables:

mm_to_km=0.00001 km/ mm

r_moon=384,000 km

d=0.1 mm = 0.1 mm * 0.00001 km/mm = 0.000001 km

Distance to moon = distance due to folds:

r_moon = d*2^n (n is number of folds)

Solve for n:

2^n=r_moon/d

n=log(r_moon/d)/log(2)

=38.4823…

=> n = 39 to reach to the moon + some.

Jim Turner

mrman| Profile February 4th, 2010 - 6:06 pmIf folding doubles the total thickness of the stack, then it is 39.

ceil(ln(3.84*10^11)/ln(2)) = 39

joe| Profile February 5th, 2010 - 11:13 amI saw a program here in Europe once that proved you could not fold any size sheet of paper more than 7 or 8 times. I tried it and it seems to be true, but I might have a go at the theortically possible answer over the weekend…

Tsopi| Profile February 5th, 2010 - 4:19 pmno possible…a paper can’t be fold over 8 times whatever the size of it

michaelc| Profile February 5th, 2010 - 8:07 pmHa Shawn! That’s funny. Are you trying to destroy the world’s trees? We have to go, go green!

Your way is probably more realistic than mine, if there is a realistic way that is. I keep picturing how in the world that it stays in orbit with the moon and on the earth at the same time.

Hey, we could save a lot of trees if we just found a 21 page book and just folded that once and be done with it.

Shawn| PUZZLE GRANDMASTER | Profile February 8th, 2010 - 8:54 amA good point, michaelc. How would the tides be affected by having the earth “connected” to the moon by a monster paper tower? How much surface area of the earth would be covered by the base of the tower? Is it bilbao’s secret agenda to destroy all life on earth????

bilbao| Profile February 11th, 2010 - 1:51 amHi Shawn, you’ll agree with me that fulfilling my secret agenda with just a sheet of paper is huge…

michaelc, Shawn, it is great that this puzzle also made your imagination work :-)

42 is the answer that I have for this puzzle. Will try to post more ‘short and sweet’ puzzles…

Hendy| Profile September 13th, 2010 - 4:13 pmBTW. Watch Mythbusters they actually folded a “piece” of paper 11 times. This was not completely one piece, but the seams were not too bad. http://en.wikipedia.org/wiki/M.....even_Folds

Be careful about the rules. If it were a common 8.5 x 11 sheet of paper, one would have a very hard time folding molecule sized folds (or slices if stacking).

The ability to fold versus the actual thickness is a topic of debate. As you can see as you fold a piece of paper (of any size) the folds themselves in addition to the paper start to bunch up. The paper loses surface area to touch another part of the paper. There are air pockets or distances of gap if in a vacuum.