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A pencil has dimensions 5E-3 m x 5E-3 m x 1.5E-2 m =3.8 E-7 m^3.
The box has volume 2E3 m^3

So 2E3/3.8E-7= about E10


Estimate 16^16

you-know-who wrote:A pencil has dimensions 5E-3 m x 5E-3 m x 1.5E-2 m =3.8 E-7 m^3.
The box has volume 2E3 m^3

So 2E3/3.8E-7= about E10


Estimate 16^16

ok, so 16 is 2^4
so (2^4)^16 = 2^64
and 2^10=10E3
so 2^60 = ~10E18
and since 2^4 is 16, i'd say it would be somewhere about E19.

Fermi is really simple event once you get the basics down and have the important facts memorized. Are there any challenging problems that someone wants to write for me with an actual answer?
Mine is: How many hydrogen atoms are in the Sun? (it seems really annoying but it's pretty simple)

DivineBbbbbeast wrote:Fermi is really simple event once you get the basics down and have the important facts memorized. Are there any challenging problems that someone wants to write for me with an actual answer?
Mine is: How many hydrogen atoms are in the Sun? (it seems really annoying but it's pretty simple)

You don't need the answer key as if you want the answer, you can just look up how close your estimations were.
From my days of Fermi, mass of sun = 2E30kg.
Assume all are H (most you can be off by is maybe 30% over so round down if you are close).
Then you have 2E33 g (1g/mol) (6E23/mol) => 57 as your answer.
You could also use the well known facts that sun's volume ~ 1E6 earth's volume, earth's mass = 6E24kg and sun's density is some fraction of earth's to find the sun's mass.

Here's a challenge: How many Helens does it take to launch enough warships to cover the entire earth with intercontinental ballistic missiles?

alecfxl wrote: Here's a challenge: How many Helens does it take to launch enough warships to cover the entire earth with intercontinental ballistic missiles?

Well this one is a little out there but after some googling (Yeah I cheated, get at me) I found that Helen "is the face that launched a thousand ships" so a milliHelen is the amount of beauty needed to launch one ship.

Then I know the surface of the earth is 5E11 square meters, and I'm guessing a ballistic missile is E1 meters long and E0 meters wide, so the area of one side of a ballistic missle is E1.
Then if I take a 1:1 ratio of missiles launched to ships launched, you get 5E10 ships, convert that to Helens, and you get 5E7, rounding up to E8.

Whats the mass of all the dandruff in America?

I thought "Is this the face that launched a thousand ships?" would be fairly well-known and was actually from a test. I just added the bit about ICBMs, which I didn't word well enough. Point was each warship (launch position) would be able to cover a very large area due to the ICBM's range, which you would estimate to be thousands of miles in radius since they are intercontinental. Then, you'd divide area of earth by the effective range (all parts of the world are close enough to sea for ICBMs).

Dandruff in US. Since you didn't state human dandruff, I'll have to take some wild guesses.
3E8 humans but add animals you are up to ~1E9 effective humans' worth of dandruff. Dandruff are dead cells and there are 1E14 cells in a human body.
Scalp size is approximately (2E-1m)^2 = 4E-2. Cell size ~ (1E-5m)^2 = 1E-10. Cells/layer is then 4E8.
Assume ~2.5 layers (total guess and to make nice numbers) of dead cells shed as dandruff => 1E9 dandruff cells/being. That's 1E18 total dandruff cells or 1E4 humans.
Humans weight 70kg so 7e5 kg. I'd round down to 5 as an answer due to dandruff being dead, dried up cells.


How many redwoods would absorb all the carbon emissions in the world?

E10 Ha Ha

How many people play with lego star wars exactly once every decade?

Please post an explanation rather than simply answering the question so people can follow your thinking.

As far as the lego star wars, I am going to assume that you mean in each decade of a person's life.

Since many kids (20%?) play with legos, but very few adults, I'll assume that 2% of adults who had star wars legos as kids still play with legos since they have kids or for other reasons.
20% of the population is ~ 1.4E9 and 2% of that number is 2.8E7.
This is the point where it comes down to almost completely guessing. I will say that 0.05% of adults who play with legos happen to exactly once per decade, giving the answer 1.4E4

Number of redwoods to absorb all CO2 emissions:

It will be difficult to estimate the amount of CO2 absorbed per tree, so I am going to try a more inventive method.

Plants are responsible for 0.25 the total C02 absorption. From a memorized list of prominent state park sizes, I know that the major redwood forest is about 1.5E5 acres ~ 4E-6 the land area of the earth, and relative to the mean, this area has 5 times the biomass, so redwoods have 2E-5 the total biomass of plants on earth and therefore CO2 absorption of all plants. I have heard a statistic that we release about 150 times the natural level of CO2, so to establish equilibrium you would need 150 times as many C02 absorbers. That gives us 150*4*5E4 = 3E7

Question:

Estimate 100!. Suggestions: use Stirling's formula (Stirling's approximation) and memorize some base 10 logarithms. (log_10(e)~0.434)

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

25!*32!*84!*45!+120!