Fermi Questions C

[Ohai]

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?

[49ers]

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?

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!).

[EastStroudsburg13]

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!).

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.

[Ohai]

EASTstroudsburg13

That's actually just what I was thinking of, rounding the factors to their nearest tens. If I ever face one, I'll probably skip it and go back to it on the test. But, like I said before, I've never seen one and didn't even know they were on any Fermi tests until just a while ago.

[Ohai]

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!).

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.

Sorry, I messed up the quote on my last post.

[unnik9]

What is some useful background information for this event?

[OldSpice]

What is some useful background information for this event?

literally everything.

[brobo]

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!).

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.

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]

What I did was round each number, so 1-4 round to 1, 5-14 round to 10, and 15-23 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 Wolfram-Alpha 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.

[Schrodingerscat]

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).