Fermi Questions C
 OldSpice
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Re: Fermi Questions C
literally everything.unnik9 wrote:What is some useful background information for this event?
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Re: Fermi Questions C
Sorry, but I'm not following what you're saying there. I might just be misreading, but can you describe it again, please? I just got put into this event for regionals but have had no practice, and I want to do as well as possible. I bought the book How Many Licks? (Or, how to estimate d*mn near anything) which is an awesome book but it has no information on those crazy math problems. Also, is there a quick way to calculate roots (not necessarily square)?EASTstroudsburg13 wrote:49ers wrote:I have been wondering about this as well. I had one question at an invitational that had three factorials multiplied by each other (something like 23!*34!*45!).quizbowl wrote:I got 4th at the Athens  42 questions, finished in 33 minutes. No unusual questions, some funky math ones (23! * 34! * 41!, I think) and a few general randomness ones. Great job to Solon, I heard you guys got 32 5's or something :O
I'd probably round the numbers in the factorial to get 23! as 1*1*1*1*10*10*10*10*10*10*10*10*10*10*20*20*20*20*20*20*20*20*20, and then combine that to get (10^10)*(10^9)*(2^9). Since 2^9 is 512, that would be a factor of 19+3=22. However, I don't really do Fermi, so I don't know if you can round like that.
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Re: Fermi Questions C
What I did was round each number, so 14 round to 1, 514 round to 10, and 1523 round to 20. Then when all of those are combined, you wind up with 10 10's and 9 20's. 10^10 is easy enough, and for 20^9 I separated it into 10^9 and 2^9. 2^9 is 512, so that combined with (10^10)(10^9) is a factor of 22. However, you'd need to do that with 34 and 41 too. I forgot that from before. If you're crunched for time, you could probably do 23+34+41=97 and call it a day.
EDIT: I made WolframAlpha compute it for me. It's a factor of 110. Sounds like a lot of work to me for the amount of time you had.
EDIT: I made WolframAlpha compute it for me. It's a factor of 110. Sounds like a lot of work to me for the amount of time you had.
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Re: Fermi Questions C
I am curious if anyone has ever seen an event supervisor bring in a physical item for teams to visually measure to use in a Fermi question? (eg the supervisor brings in a jar of jelly beans for teams to give a Fermi answer for the number of jelly beans).
Last edited by Schrodingerscat on February 6th, 2012, 4:15 pm, edited 1 time in total.
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Re: Fermi Questions C
Nope  the closest thing to that is an image on a test. They gave us an "actual size" of something, and asked us a few questions about it. But never a physical entity.Schrodingerscat wrote:I am curious if anyone has ever seen an event supervisor bring in a physical for teams to visually measure to use in a Fermi question? (eg the supervisor brings in a jar of jelly beans for teams to give a Fermi answer for the number of jelly beans).
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 OldSpice
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Re: Fermi Questions C
Was that on your regional's test?quizbowl wrote:Nope  the closest thing to that is an image on a test. They gave us an "actual size" of something, and asked us a few questions about it. But never a physical entity.Schrodingerscat wrote:I am curious if anyone has ever seen an event supervisor bring in a physical for teams to visually measure to use in a Fermi question? (eg the supervisor brings in a jar of jelly beans for teams to give a Fermi answer for the number of jelly beans).
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Re: Fermi Questions C
One question: why is it that a number beginning with a digit 5 or greater is rounded to the next power of 10, and a number beginning with a digit less than 5 is rounded down? The event asks us to estimate the answer to the nearest power of 10, so that's equivalent to estimating the log of the answer to the nearest whole number. Then shouldn't it make sense to round up when the decimal part of the log is 0.5 or greater and round down when the decimal part of the log is less than 0.5? This corresponds to rounding to the next power of 10 when the original number is greater than the square root of 10 (about 3.16) and rounding down when the original number is less than the square root of 10.
Also, regarding the factorials: somebody at my region apparently devised a REALLY ACCURATE method for estimating factorials. He claims he was able to estimate 100! within a factor of 1.4 . He's not telling anyone the details about this method though...
Also, regarding the factorials: somebody at my region apparently devised a REALLY ACCURATE method for estimating factorials. He claims he was able to estimate 100! within a factor of 1.4 . He's not telling anyone the details about this method though...
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Re: Fermi Questions C
Yeah, I can't imagine there's any good reason for that besides "it's easier". Rounding should really be determined by 0.1*sqrt(10) (0.3ish, close enough) rather than 0.5.youknowwho wrote:One question: why is it that a number beginning with a digit 5 or greater is rounded to the next power of 10, and a number beginning with a digit less than 5 is rounded down? The event asks us to estimate the answer to the nearest power of 10, so that's equivalent to estimating the log of the answer to the nearest whole number. Then shouldn't it make sense to round up when the decimal part of the log is 0.5 or greater and round down when the decimal part of the log is less than 0.5? This corresponds to rounding to the next power of 10 when the original number is greater than the square root of 10 (about 3.16) and rounding down when the original number is less than the square root of 10.
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Re: Fermi Questions C
All you really have to know for factorials is that n! is approximately (n/e)^n*sqrt(2pi*n), so taking a logarith base 10 gives a general equation for the order of magnitude:Ohai wrote:I haven't faced any factorials in any of the tournaments I've been to. They must not be that popular in the Kansas region... But just in case, how would one ideally (quickly, fairly accurately) find a factorial during a Fermi Test? Or do you just have to memorize general ones?
log_10(n!)=log_10((n/e)^n*sqrt(2pi*n))
log_10(n!)= n(log_10(n)log_10(e))+1/2(log_10(2pi)+log_10(n)
log_10(n!)=n(log_10(n)0.434) + 1/2(0.798+log_10(n))
and then just remember that log_10(5) is 0.7, so if this value gives x.7 or higher, then put down x+1, if it is x.69 or lower, then put down x. For large n this is as accurate as your memorized values for log_10(e) and log_10(2pi), so within 1%.
Similarly, you can do things for powers of 2, e, 3, etc given you have some logarithms memorized.
11^1111 would be difficult if you didn't know log_10(11) was 1.04 now we can just get log_10(11^1111) = 1111log_10(11) = 1111*1.04 = 1155.55 which gives our Fermi answer as 1155, close to the actual value of 1157.
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