EPTEmpyrean's Math Problem of the Day - Page 3
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Darki[Per]
Peru689 Posts
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Cambium
United States16368 Posts
It's 3:39 AM, and I just read Catyoul's solution, it didn't make too much sense to me. Anyway, I'll try to solve it tomorrow morning. Keep it up! | ||
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Cambium
United States16368 Posts
On December 16 2005 19:32 Catyoul wrote: Nice thread  Looks like I'm in time for day 2. Basically, we're looking for the highest power of 10 in 100000!, that is the minimum between the highest power of 2 and the highest power of 5. It's obvious that the highest power of 2 is higher, so we just have to find the highest power of 5 in 100000! To find that, we'll need to find how many multiples of 5, 5^2, 5^3,... there are between 1 and 100000, and that is simply : - 5 : 100000/5 rounded to the inferior = 20000 - 5^2 : 100000/25 = 4000 - 5^3 : 100000/125 = 800 - 5^4 : 100000/625 = 160 - 5^5 : 100000/3125 = 32 - 5^6 : 100000/5^6 = 6 - 5^7 : 100000/5^7 = 1 That makes 20000+4000+800+160+32+6+1=24999, which means we have 5^24999 in 100000!, and we have 24999 zeroes at the end of 100000! It's getting late (like I said before), but shouldn't you consider those things? - Overlapping, e.g. 5 and 25, you count them twice - You should multiply the corresponding power to the numbers (You need to subtract the numbers from each other, like I said before), so, assume they are right, they should be 20000*1 + 4000*2 +800*3 + 160*4 + 32*5 + 6*6 + 1*7 That's how many 5's you have in 100000! Since there are obviously more 2's than 5's, number of 5's represent number of 0's, plus all the multiple of 10's, give you the correct solution (like you said). | ||
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Cambium
United States16368 Posts
Then again.... it's unlikely. Anyway, GN people ^^ | ||
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matamata
United States133 Posts
On December 17 2005 00:56 Cambium wrote: It's getting late (like I said before), but shouldn't you consider those things? - Overlapping, e.g. 5 and 25, you count them twice - You should multiply the corresponding power to the numbers (You need to subtract the numbers from each other, like I said before), so, assume they are right, they should be 20000*1 + 4000*2 +800*3 + 160*4 + 32*5 + 6*6 + 1*7 That's how many 5's you have in 100000! Since there are obviously more 2's than 5's, number of 5's represent number of 0's, plus all the multiple of 10's, give you the correct solution (like you said). Every multiple of 25 is also a multiple of 5, every mult of 125 is already a mult. of 5, 25, 125 etc... so when he counted all multiples of 5, he counted all 25's once already, all 125's once as well... when he counted all 25's, he counted all 125's once again, all 625's etc... So he needs to add each successive power only once, leading to a pretty answer :D | ||
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nicetoknowyou15
Jamaica185 Posts
On December 16 2005 12:59 Jin wrote: Um... z(z^2-174) = 308 z^2-174 != 0 -14 | 1 0 -174 -308 | 0 -14 -196 -308 -------------------- gives 1 14 22 0 so you get (z-14)(z^2 +14z +22) rest is trivial I realize the syntax might not make any sense but I can't draw it out in ascii. I forget why but when dealing with qudrilaterals, once you set it equal to zero and factor it, you can get solutions for anything that is left in the parenthasese by setting THAT equal to zero So that's the answer I got ... I don't even know the meaning syntax ... like no clue what it is. Also ... there are 5 zeros 100000 | ||
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Cambium
United States16368 Posts
I thought about this property (read post before) but had trouble dealing with odd powers, now it all makes sense. Elegant, indeed. My method would've worked, save it's longer. -.-!! | ||
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User11
Canada481 Posts
answer is none  no zeros ^_^ | ||
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Return
Ivory Coast856 Posts
100000 | ||
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LastWish
2015 Posts
On December 17 2005 01:02 matamata wrote: Ohhhh.... math <3 Every multiple of 25 is also a multiple of 5, every mult of 125 is already a mult. of 5, 25, 125 etc... so when he counted all multiples of 5, he counted all 25's once already, all 125's once as well... when he counted all 25's, he counted all 125's once again, all 625's etc... So he needs to add each successive power only once, leading to a pretty answer :D Well I've done probably something wrong because it nothing pretty so far. So every second multiple of 5 is a multiple of 10, but a multiple of 10 may have more than one 0, resulting in 11111 zeroes for all multiplies of 10 in 10000!. Then as already said every multiple of 25 is a multiple of 5 and so on... Making it 13021 zeroes for all numbers ending with 5. 11111+13021=24132  | ||
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Jin
Canada439 Posts
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Empyrean
17004 Posts
I told you guys that one was easier.Anyway, day three is up. I believe this is possible, but I actually don't know how to do it myself exactly. You can use higher order curves I believe, but I've yet to do it myself. Oops, there's a hint. Get on it If no one gets it in a while maybe I'll just move on  This one may take a few days, but in the meantime I'll post some more. | ||
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Locked
United States4182 Posts
On December 17 2005 15:55 Empyrean wrote: Damn, Catyoul's good I told you guys that one was easier.Anyway, day three is up. I believe this is possible, but I actually don't know how to do it myself exactly. You can use higher order curves I believe, but I've yet to do it myself. Oops, there's a hint. Get on it If no one gets it in a while maybe I'll just move on This one may take a few days, but in the meantime I'll post some more.it didn't bold correctly you forgot the slash on the second b nice problems and solutions =O | ||
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Empyrean
17004 Posts
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LastWish
2015 Posts
Catyouls solution is eventually right. | ||
Bill307
Canada9103 Posts
On December 15 2005 19:43 Empyrean wrote: Day three: We all know it's impossible to construct a circle with the same area as a square within the Archimedean constraints (used only straight-edge and compass, and must be done in a finite number of constructions). Show how to construct a circle with the same area as a square if you remove one of those constraints. (edit: removed all the useless stuff that I wrote earlier )I'm assuming that if we disregard the rule "only use a straightedge and a compass", then we are still restricted to using "realistic" tools of some kind? Otherwise, the problem trivializes when you have at your disposal, say, a ruler with infinite precision . Anyway I'm clearly no expert on constructions, but I think there should be some restriction on the set of tools you're allowed to use if you disregard that first rule. | ||
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Catyoul
France2377 Posts
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Smurg
Australia3818 Posts
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Empyrean
17004 Posts
On December 17 2005 19:29 Catyoul wrote: I say we remove the silly "same area" constraint ! I say not And by tools, if you have a perfectly accurate straight-edge, then you know pi. Simple as that. You don't have to measure it out to construct it, I believe. | ||
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LordOfDabu
United States394 Posts
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