A Hard Math Problem

[OmgWtfItsALongNameOfDoom]

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.

[blahblah18]

Re: A Hard Math Problem

Oh I remember the AIME once upon a time, man that's years ago.

[blahblah18]

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

[Kilgamayan]

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.

[Shashakiro]

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.

[OmgWtfItsALongNameOfDoom]

Re: A Hard Math Problem

good job

[blahblah18]

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

[OmgWtfItsALongNameOfDoom]

Re: A Hard Math Problem

lol, im a freshman in high school, i got a 2 or 3 lol...

[Kilgamayan]

Re: A Hard Math Problem

Oh hey I didn't read that part.

[Doug31]

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.

[bman_006]

Re: A Hard Math Problem

[-Izzy-]

Re: A Hard Math Problem

Eheh.

[madpear]

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

[flamingspinach]

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

[Shashakiro]

Re: A Hard Math Problem

wrong =P