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#1 |
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FFR Veteran
Join Date: Apr 2006
Location: Los Angeles
Age: 29
Posts: 54
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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. |
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#2 |
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FFR Player
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Oh I remember the AIME once upon a time, man that's years ago.
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#3 |
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FFR Player
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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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#4 |
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Super Scooter Happy
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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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I watched clouds awobbly from the floor o' that kayak. Souls cross ages like clouds cross skies, an' tho' a cloud's shape nor hue nor size don't stay the same, it's still a cloud an' so is a soul. Who can say where the cloud's blowed from or who the soul'll be 'morrow? Only Sonmi the east an' the west an' the compass an' the atlas, yay, only the atlas o' clouds. |
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#5 | |
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TWO THOUZAND COMBO
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Nono, you're dividing the NUMBER of zero's by 1000.
So if there are 1124 zeros, the answer is 124.
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#6 |
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FFR Veteran
Join Date: Apr 2006
Location: Los Angeles
Age: 29
Posts: 54
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good job
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#7 |
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FFR Player
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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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#8 |
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FFR Veteran
Join Date: Apr 2006
Location: Los Angeles
Age: 29
Posts: 54
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lol, im a freshman in high school, i got a 2 or 3 lol...
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#9 |
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Super Scooter Happy
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Oh hey I didn't read that part.
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I watched clouds awobbly from the floor o' that kayak. Souls cross ages like clouds cross skies, an' tho' a cloud's shape nor hue nor size don't stay the same, it's still a cloud an' so is a soul. Who can say where the cloud's blowed from or who the soul'll be 'morrow? Only Sonmi the east an' the west an' the compass an' the atlas, yay, only the atlas o' clouds. |
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#10 |
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Falcon Paaaauuuunch!!!!!!
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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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![]() Last edited by Doug31; 04-21-2006 at 07:18 PM.. |
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#11 |
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FFR Player
Join Date: Oct 2003
Posts: 28
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#12 |
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Banned
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Eheh.
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#13 |
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FFR Player
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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 Last edited by madpear; 04-21-2006 at 06:30 PM.. |
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#14 |
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FFR Player
Join Date: Jan 2006
Posts: 269
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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 |
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#15 | |
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TWO THOUZAND COMBO
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wrong =P
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#16 |
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FFR Player
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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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#17 | |
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FFR Player
Join Date: Jan 2006
Posts: 269
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Quote:
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#18 | |||
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TWO THOUZAND COMBO
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Quote:
Quote:
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#19 | |
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FFR Player
Join Date: Jan 2006
Posts: 269
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ROFL I'm a dumbass whoops - I ****ed up the summation of the first part, that should be:
Quote:
hahaha, sorry ^^;; I'm pretty sure I did it right on the actual AIME though :P Last edited by flamingspinach; 04-22-2006 at 05:29 PM.. Reason: mod 1000 |
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#20 |
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FFR Player
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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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