Fermi Questions C

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Re: Fermi Questions C

Post by astroblue » September 5th, 2011, 5:26 pm

So does we start one or does a mod?
Either way, I think this should be the first question :P (remember, answers have to be powers of 10): 70!
Not really, a "typical" fermi question, but it requires estimation skills either way.
Also, it may be beneficial to memorize the logs of a few small prime numbers for this problem.

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Re: Fermi Questions C

Post by hmcginny » September 5th, 2011, 5:53 pm

astroblue: there is already one in the posting games section of the general chat board.
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Re: Fermi Questions C

Post by zxcvb » September 10th, 2011, 11:52 am

Are you allowed anything in with you (paper, pencil, calculator)?

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Re: Fermi Questions C

Post by Phenylethylamine » September 10th, 2011, 1:13 pm

zxcvb wrote:Are you allowed anything in with you (paper, pencil, calculator)?
I don't have the rules yet, but I'd certainly assume you're allowed a pencil... As for the calculator, I'd guess no, because you're more or less just multiplying orders of magnitude- i.e., adding.
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Re: Fermi Questions C

Post by quizbowl » September 10th, 2011, 1:52 pm

zxcvb wrote:Are you allowed anything in with you (paper, pencil, calculator)?
Just a writing utility. Nada mas.
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Re: Fermi Questions C

Post by Angstrom » September 11th, 2011, 6:11 am

This event is going to be fun. If any of you are interested in a more precise and less "silly" (by that I only meant the type of problems) form of Fermi - i.e. the original Fermi problem with the atomic bomb, and things like that, you should check out Caltech's Order of Magnitude Physics class - http://www.its.caltech.edu/~oom/ is the link. I don't think it's a good study resource for Fermi as an event, but it's fun to go through, and for me, that's more important.

And by the way, 70! is 10^106. I say this because a couple months ago, I was evaluating factorials and my calculator stopped at 66 or 67 factorial, suggesting that that's where you hit 10^100, so sqrt(67) * 68 * 69 * 70 = 70^3.5 = 343000 * sqrt(70) = 3e6.
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Re: Fermi Questions C

Post by quizbowl » September 11th, 2011, 9:24 am

Angstrom wrote:This event is going to be fun. If any of you are interested in a more precise and less "silly" (by that I only meant the type of problems) form of Fermi - i.e. the original Fermi problem with the atomic bomb, and things like that, you should check out Caltech's Order of Magnitude Physics class - http://www.its.caltech.edu/~oom/ is the link. I don't think it's a good study resource for Fermi as an event, but it's fun to go through, and for me, that's more important.

And by the way, 70! is 10^106. I say this because a couple months ago, I was evaluating factorials and my calculator stopped at 66 or 67 factorial, suggesting that that's where you hit 10^100, so sqrt(67) * 68 * 69 * 70 = 70^3.5 = 343000 * sqrt(70) = 3e6.
Actually, 70! is about 10^100.
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Re: Fermi Questions C

Post by Bogoradwee » September 12th, 2011, 6:29 am

quiz is right. I just checked with the calculator on my computer. I got 1.198e100
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Re: Fermi Questions C

Post by lllazar » October 5th, 2011, 4:19 pm

Ok so i'm a bit perplexed right now by the answer key for this test:

Test: http://scioly.org/w/images/e/e2/2008_FermiWayzata.pdf
Key: http://scioly.org/w/images/c/cc/2008_Fe ... ataAns.pdf

Look at questions 2-4. The answers are completely wrong, or I am failing epically at math.

, and thus the answer for 2 should be . Look at 3 and 4 as well, they should be and respectively....
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Re: Fermi Questions C

Post by hmcginny » October 5th, 2011, 4:33 pm

so for 2 it's actually saying 10 ^ (10^-10). 10^-10 is 1E-10, so the problem is really asking what is 10 ^(1E-10). 1E-10 is very close to 0 so your answer is 0. You just placed the parentheses incorrectly when you were doing it in your head.

3 is 10^(10^(10^-10)), 10^-10 is 1E-10, 10^(1E-10) is 1 as seen above, so the answer is 1.

4 is 10^(10^(10^(10^-10))), use the stuff from three to get down to 10^(10^1) which is 10^10 so the answer is 10.

I hope that helped. I think your only problem was your idea of where the parentheses would go.
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