Fermi Questions Marathon

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Re: Fermi Questions Marathon

Postby samm547 » March 7th, 2012, 5:00 pm

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 ?

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Re: Fermi Questions Marathon

Postby quizbowl » March 7th, 2012, 5:04 pm

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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Re: Fermi Questions Marathon

Postby samm547 » March 7th, 2012, 5:19 pm

(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

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Re: Fermi Questions Marathon

Postby samm547 » March 7th, 2012, 5:33 pm

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

Postby Schrodingerscat » March 12th, 2012, 4:42 pm

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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Re: Fermi Questions Marathon

Postby samm547 » March 14th, 2012, 5:13 pm

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?
Assumptions: Car gets 20mpg, Dallas to Chicago is 2E3 miles, Ink cartrige is 20 ml and costs 25$, 2L is 1 gallon

Then it is simple multiplication and division, giving 2E3/20*2*1000/20*25 = 2.5E5

What is the 1,000,000th prime number? (This was on my regional test) If stuck, use prime number theorem.

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Re: Fermi Questions Marathon

Postby quizbowl » March 14th, 2012, 5:23 pm

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?
Assumptions: Car gets 20mpg, Dallas to Chicago is 2E3 miles, Ink cartrige is 20 ml and costs 25$, 2L is 1 gallon

Then it is simple multiplication and division, giving 2E3/20*2*1000/20*25 = 2.5E5

What is the 1,000,000th prime number? (This was on my regional test) If stuck, use prime number theorem.
Okay, this is going to be a stream-of-consciousness problem.
good god, this sounds difficult. not really sure how to approach this mathematically. Well I remember that in the first 100 numbers, 25 or so are prime. so that's about 1/4th, right? but i feel like as you go up it'll get lower, like maybe 1/8th or so. Let's say around 1/8th then. So one million primes, wow that's a lot of primes. holy cow that's huge hehe. uhm so really going by that ratio by the 8 millionth number we should have about the lucky number somewhere by then. So I'd say 7? [/stream]

What is the density of the universe (on average) in kg/cubic light year?
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Re: Fermi Questions Marathon

Postby Schrodingerscat » March 15th, 2012, 9:05 am

I believe there are a billion galaxies averaging 1 trillion stars each, so will go with 10^21 stars, each weighing about 10^30 kg, so 10^51 kg.
The universe has a radius of I believe around 5*10^10 lys, so 100*10^30 or 10^33 cubic light years.
so thus the density is 10^18 kg/ly^3. Fermi answer 18.

What is the cost of a kilogram of Silver nitrate?

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Re: Fermi Questions Marathon

Postby quizbowl » March 18th, 2012, 7:46 am

What is the cost of a kilogram of Silver nitrate?
Ok, so silver nitrate itself is a bit expensive. I flipped through Ward's a few weeks ago and I remember that like 25 grams was about E2 dollars or so. I think. So multiply by 4E1 and you get E3 dollars (whoa, expensive!)

How many trees would it take to remove all of the CO2 produced by the US in one year?
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Re: Fermi Questions Marathon

Postby Phenylethylamine » March 18th, 2012, 7:50 am

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?
Assumptions: Car gets 20mpg, Dallas to Chicago is 2E3 miles, Ink cartrige is 20 ml and costs 25$, 2L is 1 gallon

Then it is simple multiplication and division, giving 2E3/20*2*1000/20*25 = 2.5E5
Just for the record, a gallon is actually about 4L, so your answer should be 5E5, and therefore 6 rather than 5.
What is the cost of a kilogram of Silver nitrate?
Molecular weight of silver is about 100, nitrate is 14 + 3(16) = 60ish (both slightly low), so silver nitrate is 5/8 silver by mass; let's say the cost of silver nitrate is determined entirely by the cost of the silver it contains, so you have 5/8 kilo of silver, or 6.25E-1 kg; silver is roughly $30/oz, one ounce is 28g or 2.8E-2 kg:

6.25E-1/2.8E-2 = 2.25E1; 2.25E1*3E1 = 7E2; so 3.

Edit: oops, I was too slow; someone else already answered...
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