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A Hard Math Problem
Ok, this problem was on the AIME (American invitational mathematics examination). I got in two months ago and god, it was hard... Here's a problem from it:
Let N be the number of consecutive 0's at the right end of the decimal representation of the product 1! (factorial) 2!3!4!....99!100!. Find the remainder when N is divided by 1000. I got this after quite a long time of thinking...it's a 3-digit integer. I'm in Math analysis and you should be able to solve it with Math Analysis knowledge. Good luck. |
Re: A Hard Math Problem
Oh I remember the AIME once upon a time, man that's years ago.
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Re: A Hard Math Problem
ok that's really piss easy for an AIME question, it just has to go by the fact that every # after 5! adds another 0 until you're adding 00 with each one etc etc, then divide that by 1000 is jsut to make it a 3 digit answer so it fitrs on the scantron.. too lazy to do the 5 seconds of math behind it
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Re: A Hard Math Problem
Why would there be a remainder? 100! alone contains a 100*10, which is 1000. Since multiplication and division are commutative, the two 1000s will kill each other and you'll be left with (1!2!3!...99!99!)/10, which is most certainly a whole number.
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Re: A Hard Math Problem
Nono, you're dividing the NUMBER of zero's by 1000.
So if there are 1124 zeros, the answer is 124. |
Re: A Hard Math Problem
good job
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Re: A Hard Math Problem
what'd you get on the AIME, I think my junior year of high school I got 6, which i was fairly proud of
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Re: A Hard Math Problem
lol, im a freshman in high school, i got a 2 or 3 lol...
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Re: A Hard Math Problem
Oh hey I didn't read that part.
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Re: A Hard Math Problem
Ya, I easily see how to do it. Every number 5 and up adds 1, every number 10 and up adds two, every number 15 and up adds three, every number 20 and up adds four, every number 25 and up adds six, since 25 has 2 5s in it, and so on, all the way up to 100, and add up all these.
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Re: A Hard Math Problem
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Re: A Hard Math Problem
Eheh.
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Re: A Hard Math Problem
Yea I'm just finishing taking multivariable calc along with algorithms and data structures at the U of MN IT. You want a challenging math problem, try surface integrals of vector fields like heat flux...
But for that specific problem you can just take Zeros in(1!2!...100!)modulus1000=ans |
Re: A Hard Math Problem
think of it like this - 100^1*99^2*98^3*...*1^100
All possible multiples of five there are: 100^1, 95^6, 90^11 ... 5^96 All multiples of 25 also have an extra five so count those as well: 25^76, 50^51, 75^26, 100^1 So that's (1+6+11+16+...+91+96) + (76+51+26+1) = (1 + 5 sum(i=0 to 19,i)) + (154) = 1 + 5(19)(20)/2 + 154 = 155 + 50*19 = 155 + 950 = 1105; 1105 mod 1000 = 105, which is the answer You guys forgot the squares, I think o.o But yeah this is one pretty easy AIME problem :) I'm surprised none of you actually did it yet :P |
Re: A Hard Math Problem
wrong =P
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Re: A Hard Math Problem
surface integrals are piss easy, there's no thought involved in any level of calculus, who are you kidding if that's mind-straining
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Re: A Hard Math Problem
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Re: A Hard Math Problem
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Re: A Hard Math Problem
ROFL I'm a dumbass whoops - I ****ed up the summation of the first part, that should be:
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hahaha, sorry ^^;; I'm pretty sure I did it right on the actual AIME though :P |
Re: A Hard Math Problem
blahblah apparently you never did challenging surface integrals, that or they always let you evaluate them on mathematica. when you get iterated integrals with 7th degree trig functions in them, have fun doing that by hand and telling me it is easy
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