Paper Folding exponetial y=0.1(2^x) solving problem

[MJFan]

Hi!
Here is the question:

The Paper Folding Problem was a well-known challenge to fold paper in half more than seven or eight times, using paper of any size or shape. The task was commonly known to be impossible until April 2005, when Britney Gallivan solved it.


A sheet of letter paper is about 0.1 mm thick. On the third fold it is about as thick as your fingernail. On the 7th fold it is about as thick as a notebook. If it was possible to keep folding indefinitely, how many folds would be required to end up with a thickness that surpasses the height of the CN Tower, which is 553 m?

Here is what I got:

So therefore, you would have to fold it 20 times to have a height taller than the CN Tower, since you cannot fold 19.08 times.
But when I googled the question, I kept getting answers of 23.
I was wondering if I have solved the question correctly?

 

[Ishuda]

Hi!
Here is the question:

The Paper Folding Problem was a well-known challenge to fold paper in half more than seven or eight times, using paper of any size or shape. The task was commonly known to be impossible until April 2005, when Britney Gallivan solved it.


A sheet of letter paper is about 0.1 mm thick. On the third fold it is about as thick as your fingernail. On the 7th fold it is about as thick as a notebook. If it was possible to keep folding indefinitely, how many folds would be required to end up with a thickness that surpasses the height of the CN Tower, which is 553 m?

Here is what I got:

So therefore, you would have to fold it 20 times to have a height taller than the CN Tower, since you cannot fold 19.08 times.
But when I googled the question, I kept getting answers of 23.
I was wondering if I have solved the question correctly?

Since it is 0.1 mm, the initial equation is
553 = (0.1/1000) 2^x
or
2^x = 5530000
and not
2^x = 553000
as you have.

The answer then becomes x ~ 22.3989 or the number of foldings required is 23.

 

[Ishuda]

This one is interesting:
you work for 31 consecutive days:
you're paid .01 on 1st day, .02 on 2nd day, .04 on 3rd day,
and so on...keep doubling to 31st day.
How much will you have earned?

\(\displaystyle 0.01\, *\, \underset{k=0}{\overset{k=30}{\Sigma}}\, 2^k\) = A WHOLE LOT! > 20 000 000

 

[Steven G]

Yep. $21,474,836.47 :cool:

I do not believe it is that much. Let's give it a try, you can give me a penny tonight and double it every day for 30 days and I bet you will not have even given me $10. I will message you with my paypal account info.

 

[Ishuda]

1: .01 .01
2: .02 .03
3: .04 .07
4: .08 .15
5: .16 .31
6: .32 .63
7: .64 1.27
8: 1.28 2.55
9: 2.56 5.11
10: 5.12 10.23
11: 10.24 20.47
keep going!

Ah yes, Denis. But give Jomo some $21 million and he will gladly pay off even a $10 million dollar bet.