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On June 17 2005 19:15 lightman wrote:Show nested quote +On June 17 2005 18:56 BCloud wrote:
My teacher said that if you solved it you would win a fields medal or something like that, which would be the equivalent to nobel, but i guess everyone knows the answer now --v? Oh and Andrew won nothing, and he has said he doesn't really want anything. What he sure won was a place in history as one of the greatest by solving one of the most (the most for some people) intriguing math problems of all time. Oh and by the way, do you suck your teacher's cock or something?
my teacher is a she idiot, next time read it carefully
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You have three spinners with 4 slots on each one. Assign the number 1-12 on them so that spinner A beats spinner B on average, spinner B beats spinner C on average and spinner C beats spinner A on average.
(by beat I mean has a higher average)
Show a proof as well, dont simply list numbers. The proof will not be an equation, but it will be a logical set.
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You mean that when you spin the spinners (lets say its A and B), the # that comes up on spinner A is greater than the # on spinner B more than half the time?
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Calgary25986 Posts
Well I had it done but I don't have "proof". I don't really know what you expect from that... >_<
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yes. exactly right. I mean, "on average" spinner A beats B, B beats C, and C beats A.
Basically, the majority of the time A >B>C>A. All you need to show is a scenario where this happens more than 50% of the time for each one.
PM me the number sets chill.
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Catyoul
France2377 Posts
On June 17 2005 19:06 lightman wrote:
So far it looks like BigBalls is good at this I congratulate him for both being smart and intereseted, and for being passionate and hunger to this. But also I may say so far I sense his talent hasn't been challenged yet, so I'll start to warm him up (or anyone else that accepts the challenge). It's easy, requires some reasoning and math and that's it:
Given an equilateral triangle with a side that lenghts L. Each one of the three sides is divided into n+1 equal parts, and for each point that divides a side, you draw n parallel lines to each one of the three sides.
Calculate the total number of triangles obtained, this is, the total number of equilateral triangles that will have by side-lenght L/n , 2L/n , .....L.
Just got back home, 7am, I'm dead tired so I hope this will make sense 
This problem looks cool (had already seen all the other problems of the thread). I'm pretty sure you mean L/(n+1), 2L/(n+1), etc. for the lengths, anyway I'll use n as your n+1 as defined initially or your n as defined by the side-lenghts.
Let k(p,m) be the number of triangles of side-lengths p*L/m when n=m.
n=1, k(1,1) = 1 (number of triangles of length L at n=1)
n=2, k(2,2) = 1, k(1,2) = 3+1
n=3, k(3,3) = 1, k(2,3) = 3, k(3,3) = 5+3+1
and we could go on...
More generally, when we go one step further at n, we could see it as enlarging the triangle at (n-1) by one line, so we just have to count how many triangles of the good length we are adding. First we are adding new triangles which base belongs to the new bigger base (new triangles facing er... upside up ?) and we'll have how many... we'll have n-(p-1) for k(p,n). We shouldn't forget the new triangles which don't have their base but just one point on the new enlarged triangle base (new triangles facing upside down) and we have something like (n-p)-(p-1) for k(p,n) that is n+1-2p, or more exactly the max of either that or 0 if we don't have any such triangle, which can happen.
Soooooo, we should have something like k(p,n) = k(p,n-1) + n+1-p + max(n+1-2p,0) and with that we can find all of them. Maybe we could find an elegant formulat, but that will be left as dessert for whoever wants to.
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Catyoul
France2377 Posts
What's a spinner ?
edit : here's another sequence (haven't found xeo's yet -_-), though I'm pretty sure I must have posted it already :
10, 11, 20, 31, 52, 121, 200
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On June 17 2005 22:51 Catyoul wrote:
What's a spinner ?
edit : here's another sequence (haven't found xeo's yet -_-), though I'm pretty sure I must have posted it already :
10, 11, 20, 31, 52, 121, 200
314,512(hah! easy one),851,1228
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You have ten stacks of coins, and each one has 10 coins in it. However, one of the stacks consists entirely of counterfeits, which each weigh one gram more than a standard coin. You know the weight of a standard coin. You have a scale that will tell you the exact weight of whatever you put on it, but for the sake of having an interesting puzzle you may only use this scale once.
How do you determine which stack is counterfeit?
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On June 17 2005 13:39 loloko2 wrote:
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
OK HERE IS ANOTHER ONE, I NEED HELP ON THIS, BECAUSE IT WAS FOR HOMEWORK A FEW DAYS AGO AND I NEED ALOT OF HELP , PLEASE...
Why does the equation X^n + Y^n = Z^n it doesnt have integer solutions if n > = 3 .
Find me a solution please. Or if it cant be done then explain why you cant?.
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
X = 0, Y = Z ;
X = Z, Y = 0;
X = -Y, Z = 0, n is odd;
It seems that there's plenty of integer solutions. 
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On June 17 2005 23:26 LordOfDabu wrote:Show nested quote +On June 17 2005 13:39 loloko2 wrote:
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP
OK HERE IS ANOTHER ONE, I NEED HELP ON THIS, BECAUSE IT WAS FOR HOMEWORK A FEW DAYS AGO AND I NEED ALOT OF HELP , PLEASE...
Why does the equation X^n + Y^n = Z^n it doesnt have integer solutions if n > = 3 .
Find me a solution please. Or if it cant be done then explain why you cant?.
PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP X = 0, Y = Z ; X = Z, Y = 0; X = -Y, Z = 0, n is odd; It seems that there's plenty of integer solutions.
Umm no...
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Um, yes. Checking them yourself should be pretty simple.
With that said, the correct statement of the theorem is supposed to be that it has no non-zero integer solutions.
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Good to see Catyoul and BigBalls back tearing shit up with these math puzzles again. Couple new faces as well. Fun times with that boy/girl percentage problem where I made an ass of myself for at least a good 10 pages. I've been helping out with some of the contest work at my local high school, this was the last question off of the Euclid (grade 12 contest). Haven't taken the time to work out part c) yet, the first two are easy.
Consider a list of the first n positive integers (the problem was stated with opening/closing lockers but that's not needed). Proceed in this fashion, beginning with 1, remove every second number from the list (i.e. 2, 4, 6, ...) until you reach the end. Then of those numbers that remain starting with n and going backwards, again remove every second number. Again repeat this procedure going back up and down the list until there is only one number left. Let f(n) be this number.
Ex. n = 15
We remove
2 4 6 8 10 12 14
13 9 5 1
7 15
3
This gives us f(15) = 11
a) f(50) = ? (Grunt work if you want)
b) Show there is no n such that f(n) = 2005 (Easy)
c) Show there are infinitely many n such that f(n) = f(2005)
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On June 17 2005 23:22 LordOfDabu wrote:
You have ten stacks of coins, and each one has 10 coins in it. However, one of the stacks consists entirely of counterfeits, which each weigh one gram more than a standard coin. You know the weight of a standard coin. You have a scale that will tell you the exact weight of whatever you put on it, but for the sake of having an interesting puzzle you may only use this scale once.
How do you determine which stack is counterfeit?
Take 1 coin from stack 1, 2 coins from stack 2, etc. Weigh the 55 coins.
If the weight of a genuine coin is y, the correct weight should be 55y.
Then deduce which stack is the counterfeit based on how much you are off by.
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On June 17 2005 23:11 LTT wrote:Show nested quote +On June 17 2005 22:51 Catyoul wrote:
What's a spinner ?
edit : here's another sequence (haven't found xeo's yet -_-), though I'm pretty sure I must have posted it already :
10, 11, 20, 31, 52, 121, 200 314,512(hah! easy one),851,1228
is that 2^1 base 2, 2^2 base 3, 2^3 base 4, etc?
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On June 18 2005 00:41 gg2w wrote:
Good to see Catyoul and BigBalls back tearing shit up with these math puzzles again. Couple new faces as well. Fun times with that boy/girl percentage problem where I made an ass of myself for at least a good 10 pages. I've been helping out with some of the contest work at my local high school, this was the last question off of the Euclid (grade 12 contest). Haven't taken the time to work out part c) yet, the first two are easy.
Consider a list of the first n positive integers (the problem was stated with opening/closing lockers but that's not needed). Proceed in this fashion, beginning with 1, remove every second number from the list (i.e. 2, 4, 6, ...) until you reach the end. Then of those numbers that remain starting with n and going backwards, again remove every second number. Again repeat this procedure going back up and down the list until there is only one number left. Let f(n) be this number.
Ex. n = 15
We remove
2 4 6 8 10 12 14
13 9 5 1
7 15
3
This gives us f(15) = 11
a) f(50) = ? (Grunt work if you want)
b) Show there is no n such that f(n) = 2005 (Easy)
c) Show there are infinitely many n such that f(n) = f(2005)
a. 13579 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
49 45 41 37 33 29 25 21 17 13 9 5 1
1 9 17 25 33 41 49
49 33 17 1
1 33
33
f(50)=33
b. Basically, look at the pattern. First time through, all even numbers are removed. Second time through, if n = 3mod4 or 0mod4, then all 1mod4 are removed, if n=2mod4 or 1mod4, all 3mod4 are removed. 2005 is 1 mod4, so we want n to be 2mod4 or 1mod4. So, all we have left are numbers 1mod4. 1 5 9 etc
Third time through, all 5mod8 are eliminated. 2005 is 5mod8, thus f(n) cannot be 2005
c. ill get back to this, gonna shower
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i have a thought on c!
n=1, f(n)=1
n=2, f(n) = 1
n=3, f(n) = 3
n=4, f(n) = 3
n=5, f(n) = 1
n=6, f(n) = 1
n=7, f(n) = 3
n=8, f(n) = 3
n=9, f(n) = 9
n=10, f(n)=9
n=11, f(n) = 11
n=12, f(n)=11
n=13, f(n)=9
n=15, f(n) = 11
n=17, f(n)= 1
n=19, f(n)= 3
1,5,17 are the same
3,7,19 are the same
Looks like n+2^2, n+2^4, etc are the same
so just glancing, im guessing f(n) = f(n+2^(2y)) for all y>=1
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a proof i dont see for it, hmmmmmmmm
maybe catyoul or Ltt will clean up my mess 
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On June 18 2005 00:24 LordOfDabu wrote:
Um, yes. Checking them yourself should be pretty simple.
With that said, the correct statement of the theorem is supposed to be that it has no non-zero integer solutions.
Zero isn't an integer.
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lordofdabu is right
zero is an integer, youre thinking of natural numbers malmis
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EPT
