## Fermi Questions Marathon

quizbowl
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### Re: Fermi Questions Marathon

hmcginny wrote:100! is 158, thats a value that I have memorized to help with ridiculous factorial questions.

25!*32!*84!*45!+120!
Ok, just looking at that, the 120! is a drop in the bucket compared to the rest of the equation, so forget that part.
A good thing to know is to memorize the factorial for every multiple of ten to extrapolate info.
20! = 2.5E18 and 30! = 2.7E32, so 25! should be right around the middle, like E25 or something.
30! = 2.7E32 so 32! should be around E35
80! = 7E118 and 90!= 1E138, so 84! would be near the middle but leaning a bit towards the 80! side, I'd say around E127
40! = 8E47 and 50! = 3E64, so 45! should be smack dab in the middle, like E56.
25+35+127+56 = E243.

If Robotman had a penny for every emoticon used on this site since its inception and he stacked them into a giant tower, how many light years long would it be?
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hmcginny
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### Re: Fermi Questions Marathon

website has been around for E1 years, i would say that each year there are around 3E4 posts, so thats 3E5 posts since the inception of the website. I would say 5 emoticons, so 5 pennys, each one being ~1 millimeter in thickness, so 5 mm high tower, or 2 m or -14 light years.

How many electrons are there in the atlantic ocean?
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samm547
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### Re: Fermi Questions Marathon

100!:

n! ~ sqrt(2pi*n)*n^n*e^-n) By Stirling
logn! = log(sqrt(6.5*100)+log(100^100) + log(e^-100)
= 1 + 1/2log(6.5) +100*log(100) -100loge
= 1 +0.4 +200 - 43.4 = 200 - 42 = 158

Number of e- in atlantic. Assume average depth is 4000m, atlantic covers 20% of earth, R_(+) = 6E6m, ocean is 100% water, therefore density is 1gm/cm^3

Surface area of atlantic is 4*0.2*pi*6E12 = 8E13m^2, depth of 4E3 gives volume = 2.5E17m^3. 1g/cm^3 gives mass of ocean to be 2.5E23g. mass of one mol of water is 1+1+16g = 18 g. This gives the ocean 1.3E22 mol H2O or 7E45 molecules H20, there are 10 electrons per molecule, giving 7E46.

What is the probability that a monkey given sufficient rest and food would type Hamlet on his first try, assuming he is typing randomly?

quizbowl
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### Re: Fermi Questions Marathon

samm547 wrote: What is the probability that a monkey given sufficient rest and food would type Hamlet on his first try, assuming he is typing randomly?
Haha, oh wow, I know this one from the top of my head! My sixth grade teacher made us try and figure it out and naive-old-me said that any old monkey could do it but in all honesty, the answer is pretty insane, something like -183,946. I really hope that nationals asks this question or something and then takes a picture of every competitor's face as they read it!

Imagine you took every single post in "Your Daily Random Comment" and copied all of the text, putting all of it into size 12 and Times New Roman. Assuming that you entered after every post and you omitted all emoticons and images, how many pounds of ink would you need to print all of it?
2010: 5th in NYS
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2012: 3rd in NYS
<quizbowl> ey kid ya want some shortbread
<EASTstroudsburg13> I don't know why, but I just can't bring myself to delete this post.

eta150
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### Re: Fermi Questions Marathon

quizbowl wrote:
samm547 wrote: What is the probability that a monkey given sufficient rest and food would type Hamlet on his first try, assuming he is typing randomly?
Haha, oh wow, I know this one from the top of my head! My sixth grade teacher made us try and figure it out and naive-old-me said that any old monkey could do it but in all honesty, the answer is pretty insane, something like -183,946. I really hope that nationals asks this question or something and then takes a picture of every competitor's face as they read it!

Imagine you took every single post in "Your Daily Random Comment" and copied all of the text, putting all of it into size 12 and Times New Roman. Assuming that you entered after every post and you omitted all emoticons and images, how many pounds of ink would you need to print all of it?
Well...
There are about 3e4 words, so estimating 6 letters per word (accounting for spaces and characters), 2e5 letters. At around 6e1 possibilities per character (accounting for punctuation and capitalized letters), it would be about 1e-7.
-7
For the next one:
about 1e4 posts, assuming 20 words per post, 5 letters per word, we have 1e6 letters. Printed at a thickness of 1e-2 cm (maybe?), with an area of ~1 cm * 5e-2 cm, so a volume of 5e-4 cm^3 per letter, x1e6 letters gives 5e2 mL, so 5e-1 L, assuming similar (or slightly higher) density to water, 5e-1 kg, so 1e0 lb

Here's my question:
How many orangutan(s?) could you fit inside of the five tallest man-made structures on earth?

Also, for extra credit, how long would it take a swallow to fly a coconut through those buildings' combined heights (assuming they were placed end to end, on their sides) in seconds
#ACESWILD

samm547
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### Re: Fermi Questions Marathon

We assume an average height of 1750, 1500 of which are fillable feet, dimensions 15*1*1, and that Orangutans achieve maximum compression at the volume of a 3 ft by 2 ft by 2 ft cube.

That gives us total volume of 5*1.2E2^2*1500 ft = 6E4*1.5E3 = 9E7 ft^3, and 12 ft ^3 for orangutans gives 9E7/12 = 7.5E6 orangutans.

The extra credit question is bogus, he didn't specify African or European swallow.

How many digits are in the largest known prime, 2^43,112,609 - 1 ?

quizbowl
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### Re: Fermi Questions Marathon

samm547 wrote:We assume an average height of 1750, 1500 of which are fillable feet, dimensions 15*1*1, and that Orangutans achieve maximum compression at the volume of a 3 ft by 2 ft by 2 ft cube.

That gives us total volume of 5*1.2E2^2*1500 ft = 6E4*1.5E3 = 9E7 ft^3, and 12 ft ^3 for orangutans gives 9E7/12 = 7.5E6 orangutans.

The extra credit question is bogus, he didn't specify African or European swallow.

How many digits are in the largest known prime, 2^43,112,609 - 1 ?
2^10=10^3
Round 43,112,609 up a bit and divide by 10 and get 4,311,261 * 3 = 12933783.

(e^(pi^(gamma^(phi))))
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samm547
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### Re: Fermi Questions Marathon

quizbowl wrote: (e^(pi^(gamma^(phi))))
I assume my gamma you mean the E-M one,

We fill in approximate values, 2.7^3.1^0.58^1.6

estimate 0.58^1.6 = 0.45 because 0.6^2 0.36 and 0.6^1 = 0.6

3.1^0.45 ~ sqrt(pi) = 1.77 (memorized for probability theory), rounding down I'll take 1.6

2.7^1.6 is going to be slightly larger than 2.5^1.5 and sqrt(2.5) ~ 1.6 so I am going to say 5E0

samm547
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### Re: Fermi Questions Marathon

If we build a data storage device that was the size of the moon, with our present maximum storage capacity density, how many seconds of 1080p video could it hold?

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### Re: Fermi Questions Marathon

samm547 wrote:If we build a data storage device that was the size of the moon, with our present maximum storage capacity density, how many seconds of 1080p video could it hold?
I will guess that a hard drive would be a better indicator of data density than a flash drive. I believe the best hard drives have a current capacity of around 10^13 bytes (10 terrabytes) and are probably 5x5x2 or 50cm^3 Thus you can store 0.5E12 or 5*10^11 bytes in one centimeter. Actually, I will round that up to a terrabyte per cubic centimeter, as even hard drives have considerable infrastructure. That would be 10^18/cubic meter or 10^24 bytes per cubic kilometer. The earth has a radius of 6000km, and I know the moon is a fraction, so I will guess about 2000km, thus it has a volume of 4*8*10^9 km^3 or 2*10^10 cubic kilometers. Therefore I will assume the moon can hold 10^34 bytes of data I am not completely sure about how much data per second 1080 is, but I will guess 10mb/s or 10^7/s, so thus the moon can hold 10^27 seconds of video.

If cars were to run on the ink out of printer cartridges with the same "ink efficiency" as gas efficiency, how much would you spend on ink to drive from Dallas, TX to Chicago, IL?

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