Math Puzzles

On June 17 2005 13:20 Pistachio wrote:
Here's a better one:

Three woman go to their favorite restaurant and each order some food that costs £30 overall. So they each give £10 to the waiter who then gives it to the manager. The manager however decides to give £5 back to the women as a good will gesture since the women have been such loyal customers over the years. Now the waiter thinks he can make some money out of this so he secretely takes £2 for himself and only gives £1 back to each women.

This means each woman has payed £9 each, having originally payed £10 and gotten £1 back.
The waiter has £2 thus the overall money payed is 3x9 + 2 = £29.
So where'd the extra pound go?

On June 17 2005 13:15 LTT wrote:
It's not too hard. Think about it. All numbers with an even number of factors will be off, those with an odd number will be on. All primes will be off. 1 will be on.

As a result, only square numbers will be on. 1,4,9,16,25, etc. etc.

On June 17 2005 13:23 BigBalls wrote:
What is the next number in this sequence....

1,2,4,8,16,31

On June 17 2005 13:29 LTT wrote:


There are a few that could be. You are most likely thinking about the maximum number of regions formed in a circle by connecting straight lines between n points on the circle. In which case it would be, 57,99,163

On June 17 2005 13:33 BigBalls wrote:
lol i just noticed it was renamed capslock topic omfg ahahaha

someone make it math riddles or something

On June 17 2005 13:34 Jathin wrote:
LTT is probably one of those dudes that knows 1,000,000,000 useless things no one cares about =p

On June 17 2005 13:34 Jathin wrote:
LTT is probably one of those dudes that knows 1,000,000,000 useless things no one cares about =p

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

On June 17 2005 13:37 Xeofreestyler wrote:
Bigballs, its
13112221

On June 17 2005 13:41 LTT wrote:


Give Fermat a ring and ask him about his last theorem. Perhaps he will let you see the margin of his book? ... >.>

On June 17 2005 13:40 LTT wrote:
I've got one for you. Let's see if I can get this one right.

5 pirates find 100 pieces of gold and decide to split the loot. Being that they are bloodthirsty, greedy, yet rediculously rational, they come up with the following plan. (It might help if you imagine them in a line). The first pirate makes a proposal for how the loot will be split among them. If the proposal gets at least 50% of the votes, it is passed and is executed. If it fails to get 50% of the vote, the proposer is executed and the next pirate in line makes his proposal. What proposal should the first pirate make to recieve the maximum benefit possible?

On June 17 2005 13:44 -_- wrote:


Errr, he proposes that the first three pirates "in line" get 1/3 of the loot? Maybe he gets the extra coin or they spend it on strippers or something?

On June 17 2005 13:43 7op wrote:

one 1, one 3, two 1, three 2, one 1

1113213211

On June 17 2005 13:45 MPXMX wrote:
LTT, you don't buy Andrew Wiles' proof?

On June 17 2005 13:46 -_- wrote:
one 1
two 1's
one 2 and one 1
one 1 one 2 and two 1's
three 1's two 2's and one 1.

they're wrong and I'm the pirate who gets the extra coin!!!

edit: Or at least mine's sexier bitches!

On June 17 2005 13:46 LTT wrote:


No, he can do much better than that. Try thinking about it another way.

On June 17 2005 13:37 Xeofreestyler wrote:
Bigballs, its
13112221

On June 17 2005 13:43 loloko2 wrote:


Why do you have to ruin my joke to the teamliquidians T__T i hate you ^_^

On June 17 2005 13:49 Xeofreestyler wrote:


Err, wtf no?
312211 ->
one 3 one 1 two 2 two 1 -> 13112221
then that one would be one 1 one 3 two 1 three 2 one 1 -> 1113213211

Kthxbai.

On June 17 2005 13:49 Xeofreestyler wrote:


Err, wtf no?
312211 ->
one 3 one 1 two 2 two 1 -> 13112221
then that one would be one 1 one 3 two 1 three 2 one 1 -> 1113213211

Kthxbai.

On June 17 2005 13:53 BigBalls wrote:
if it was all split equally, each pirate would get 20%
if the first pirate were gone, then it were split equally, each would get 25%

50-1, 25+1, 25, 0,0

On June 17 2005 13:56 MPXMX wrote:
Must the first pirate propose specific proportions of splitting the loot or can he say things like "2nd pirate can take as much as he wants as long as 2/3 of the rest of you agree with it" ?

On June 17 2005 13:58 BigBalls wrote:
2 pirates
fourth pirate 100, last pirate 0 - passes

3 pirates
third pirate 99, fourth pirate 0, last pirate 1 - improvement

4 pirates
second pirate 99, third pirate 0, fourth pirate 0, last pirate 1

5 pirates
first pirate 98, second pirate 0, third pirate 1, fourth pirate 0, last pirate 1

On June 17 2005 13:20 Pistachio wrote:
Here's a better one:

Three woman go to their favorite restaurant and each order some food that costs £30 overall. So they each give £10 to the waiter who then gives it to the manager. The manager however decides to give £5 back to the women as a good will gesture since the women have been such loyal customers over the years. Now the waiter thinks he can make some money out of this so he secretely takes £2 for himself and only gives £1 back to each women.

This means each woman has payed £9 each, having originally payed £10 and gotten £1 back.
The waiter has £2 thus the overall money payed is 3x9 + 2 = £29.
So where'd the extra pound go?

On June 17 2005 13:20 Pistachio wrote:
Here's a better one:

Three woman go to their favorite restaurant and each order some food that costs £30 overall. So they each give £10 to the waiter who then gives it to the manager. The manager however decides to give £5 back to the women as a good will gesture since the women have been such loyal customers over the years. Now the waiter thinks he can make some money out of this so he secretely takes £2 for himself and only gives £1 back to each women.

This means each woman has payed £9 each, having originally payed £10 and gotten £1 back.
The waiter has £2 thus the overall money payed is 3x9 + 2 = £29.
So where'd the extra pound go?

On June 17 2005 13:52 -_- wrote:


-_-, do you see the three before the 1? 3 ones. do you see the 2 before the 2? Two 2's. Do you see the 1 before the 1? One one

KTHXBAIBITCH

On June 17 2005 14:03 LTT wrote:
Because if he doesn't agree, he is going to end up with nothing, and 1 coin is better than nothing.

On June 17 2005 14:08 Xeofreestyler wrote:
If you guys want another one;

1
2
10
140
5740
-Next element= ?

On June 17 2005 14:10 LTT wrote:
Wasn't that a Putnam problem? I remember seeing/doing it before.

Pacifist: Parity. (N-1) + 1/2 saved on average

On June 17 2005 14:18 Polis wrote:
You got 12 ball's, one of the weight less OR more then the others. You have balance and you can use it 3 times.

Find the diferent ball.

On June 17 2005 14:22 Xeofreestyler wrote:
Nicely done bigballs, are you on hackquest?

Is anyone able to decipher this?

34 - ?
24 - 10
16 - 7
8 - 5
4 - ?

On June 17 2005 14:18 Polis wrote:
You got 12 ball's, one of the weight less OR more then the others. You have balance and you can use it 3 times.

Find the diferent ball.

On June 17 2005 14:22 baal wrote:


Easy, you put half and half, 6 remaining, half and half 3 remaining, then you put only 2, if they weight the same is the other one.

On June 17 2005 14:10 Pacifist wrote:
i have one:

There are sixteen prisoners on death row, who are about to be executed the next day.
The warden tells the prisoners that on the day of their execution, either a white hat or a black hat will be placed on their heads. They will stand in a single file line so that the person at the front can't see any hats, the person second in line can see the first person's hat, the person third in line can see the first two people's hats, etc, and the last person in line can see everyone's hat but his own. (Nobody can see their own hat.) The warden will then, starting from the back of the line, ask each prisoner what color the hat they're wearing is. The prisoners are given the night to come up with a plan. Find the plan that guarantees the freedom of the maximum number of prisoners.

On June 17 2005 14:22 baal wrote:


Easy, you put half and half, 6 remaining, half and half 3 remaining, then you put only 2, if they weight the same is the other one.

On June 17 2005 14:27 wasted wrote:
the eyes riddle seems pretty difficult.. i'll give it a shot, i doubt it's correct, but i need a hint if im thinking in the right direction

Blue eyes present during the meeting: 125
Are there more blue or brown eyes in the meeting? same amount
Number of nights in which suicides happened: 250
Which mathematical concept made them kill themselves? statistical equal distribution

basic idea is, that you know br=num_of_browneyes, bl=num_of_blueeyes, and you assume that br=bl. your own eyecolor is x=unknown, so if it's like br=bl-1 situation, you "know" that you are bl, and kill yourself. the next night for the next person to think about it, it's bl=br, but you know what the eyecolor of the person who killed himself was. so you know that you must belong to the other group and kill yourself. da capo

On June 17 2005 14:03 LTT wrote:
Because if he doesn't agree, he is going to end up with nothing, and 1 coin is better than nothing.

On June 17 2005 14:34 Xeofreestyler wrote:


No, this isnt the right way =)
But I'll give you a hint: The mathematical concept is called induction. With other words: Proving something by proving the opposide is wrong.
If you cant find it after this I'll post the answer, its quite hard

On June 17 2005 14:35 MPXMX wrote:


if he doesn't agree, the first pirate gets executed and they can work for a better solution
whoever gets a very small share will be unhappy with the first guy getting more and vote against... as long as the first guy gets more than others, they logically shouldn't vote for it
I don't really see what perspective you are considering

On June 17 2005 14:36 BigBalls wrote:


that's contradiction

induction is when you build up for all numbers by showing a connection between n and n+1

On June 17 2005 14:34 Xeofreestyler wrote:

No, this isnt the right way =)
But I'll give you a hint: The mathematical concept is called induction. With other words: Proving something by proving the opposide is wrong.
If you cant find it after this I'll post the answer, its quite hard

On June 17 2005 14:09 BigBalls wrote:
pretty easy problem, prove answer is correct



Basketball player S's team statistician keeps track of the number, S(n) of successfull free throws S has made in his first n attempts of the season. Early in the season, S(n) was less than 80% of n but by the end of the season, S(n) was more than 80%. Is it necessarily true that at one point in the season S(n) was exactly 80% of n?

On June 17 2005 14:25 One Page Memory wrote:
And one old, not very difficult, but as far as I remember Newton gave a wrong answer first. Here it is:
There is a lake, a boat in it, and a stone in the boat. What will happen if the the stone is taken from the boat and put in the lake?
a) the level of the lake increases;
b) the level of the lake decreases;
c) the level of the lake stays the same;

On June 17 2005 14:27 BigBalls wrote:
There are 4 people who are trying to cross a bridge. One person walks across at 1 second, one at 2, one at 5, one at 10.

The bridge has holes in it, so each trip a flashlight must go across. Also, only two people may cross at once. Finally, a trip takes as long as it takes the slowest person to travel across.

Find a way to get everyone across in less than 18 seconds.

On June 17 2005 15:07 wasted wrote:


i assume this means:

there can only go 2 over at the same time, since the flashlight has to be returned, thus the following sequence must happen

side A <-> side B
4 <-> 0 // 2 from A to B = pass 1
2 <-> 2 // 1 from B to A = pass 2
3 <-> 1 // 2 from A to B = pass 3
1 <-> 3 // 1 from B to A = pass 4
2 <-> 2 // 2 from A to B = pass 5
0 <-> 4 // done

so, 5 passes and only 18 secs

5 and 10 can't go together, since one of them had to go back, which would kill the deadline.
if they have to go seperate, their trips will last fifteen secs
since you have to go back at least two times, you use at least another two secs, if you use 1 as the one who goes back. that leaves pass5, which will at least cost two seconds

that sums up to 19.

i'm curious how to do it quicker.

On June 17 2005 14:53 Xeofreestyler wrote:
Bigballs,rage, chill. Thats just the definition I heard for induction and it seemed to be the correct answer for the riddle too... Anyway;

They know they are mixed, with blue-eyed people and brown-eyed people, so there must be at least one of each. When the spaceship crashed, from the math book they learned about mathematical induction. Therefore, they knew that if there was only one brown-eyed person, the brown-eyed person would find out cos* eye color from the fact that everybody else had blue eyes, and kill cosself. If there were two brown-eyed people, each would expect the other to kill themself, and, upon finding the other alive the next morning, realize that the other person saw someone with brown eyes, which must be them, and both would die the second night. If there were three brown-eyed people, each would assume the situation in the previous sentence was taking place, until they saw that the other two were alive on the third day, at which time they would realize that there was another brown-eyed person, who must be themself, so all three would kill themselves on the third night. And so on. Since it took until the 250th night, 250 brown-eyed people killed themselves. Upon awakening the next morning and finding all of the brown-eyed people dead, the blue-eyed people would know that they had blue eyes and kill themselves the next night, leaving the planet unpopulated in just two nights of suicides.

Number of blue eyes? -249
Number of nights in which suicides happened after the meeting: -2
Are there more blue or brown eyes in the meeting? -brown
Which mathematical concept made them kill themselves? -Mathematical induction

On June 17 2005 13:21 LTT wrote:


Old One. Order of Operations.

On June 17 2005 16:27 Lord)Lw( wrote:
This one shouldn't be to hard for any of you but might as well post it.

A Robot Gazelle start at a city in Uganda and travels at a speed of 10 miles per hour toward Kenya. After reaching Kenya the Gazelle returns, traveling over exactly the same distance, at only 2 miles per hour. What is the gazelles average speed over the entire Journy? (assume that the Gazelle turned around instantly once it reached Kenya.)

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

On June 17 2005 18:20 lightman wrote:


Well, it took more than 200 years, until Andrew Wiles (a Mathematician from Princetown) through 1987-1994 came up to solve that problem.

The answer loloko2, X^n + Y^n = Z^n doesn't have integer solutions if n > = 3, is because that all eliptical cuvers are modular, a mathematical conjecture enounces by Tagimura Siyama during the late 1940s.

Oh and by the way, that problem is aka "Fermat's Last Theorem"

On June 17 2005 15:16 BigBalls wrote:


be clever

5 and 10 can't go together, since one of them had to go back, which would kill the deadline.

On June 17 2005 14:29 EvilTeletubby wrote:


Well, you could do something like, first guy says the color of the hat in front of him, so the next guy knows what to say.... and then depending upon how the guy says the color of his hat (ie, says it softly or loudly, quickly vs quiet, etc.), the in front of him would know if his hat is the same color as the previous. This would guarantee everyone but the first prisoner would live. He has 50/50 shot. ^_^

On June 17 2005 19:01 BCloud wrote:
My teacher won national math olympics these would be piece of cake for her

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?

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.

On June 17 2005 19:15 lightman wrote:


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?

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.

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

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

On June 17 2005 23:26 LordOfDabu wrote:


X = 0, Y = Z ;

X = Z, Y = 0;

X = -Y, Z = 0, n is odd;

It seems that there's plenty of integer solutions.

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?

On June 17 2005 23:11 LTT wrote:


314,512(hah! easy one),851,1228

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)

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.

On June 18 2005 07:01 lightman wrote:
The triangle problem is still unanswered

On June 18 2005 00:45 gg2w wrote:


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.

On June 18 2005 19:36 BigBalls wrote:
sorrow eyes, its not that hard even.


all lights that are hit twice are on, hit 4 times, hit 6 times.

so powers of 2, 4, 6, etc

however, powers of 4,6 are subsets of powers of 2. thus, all powers of 2 are on

On June 18 2005 19:45 BigBalls wrote:
i mean after the initial hit when theyre all on.

so 1 is hit 0 times after that

the ones that are on after that are ones that are hit an even number of times. this means they are even powers,

so they are x^(2y) = (x^y)^2, so all square numbers stay on

On June 18 2005 08:40 Catyoul wrote:

I think I've pretty much answered it. Now coming back to it with a good night (or should i say day) sleep, it's pretty straightforward to find a direct formula for each element.

k(p,n) = k(p,n-1) + (n+1-p) + max(n+1-2p,0)
= k(p,n-2) + (n-p) + max(n-2p,0) + (n+1-p) + max(n+1-2p,0)
= k(p,p-1) + (p+1-p) + ... + (2p-1+1-p) + (2p+1-p) + (2p+1-2p) + .. + (n+1-p) +(n+1-2p)
To get from 2nd to 3rd line, I've just went back step by step, k(p,p-1) is of course 0 so everything is known in that line. The separation at n=2p-1 comes from the fact that it's the point where n+1-2p gets >= 0 so we have to include the second recurrence term. Actually we have 2 cases I should have separated earlier.

Either n+1-2p >= 0 or not. If n >= 2p-1, we have the sum written above. If n <= 2p-1, the max(n+1-2p,0) is always 0, so it actually simplifies to :
k(p,n) = (p+1-p) + ... + (n+1-p)

So, we have 2 cases.
1. n <= 2p-1 :
k(p,n) = sum(k+1-p) for k=p -> n
= (n+1-p)(n+2-p)/2

2. n >= 2p-1 :
k(p,n) = sum(k+1-p) for k=p -> 2p-1 + sum(2k+2-3) for k=2p -> n
= p(p+1)/2 + (n+1-2p)(n+2-p)

Tadaaaaaaa.

On June 18 2005 21:05 decafchicken wrote:
Found these math problems on google: enjoy
1. One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there?

2. y = log x
If y = 10, then what is x?

3. 10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?

4. If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y,
then x = ?

5. What place in this world can have their temperatures Fahrenheit and Celsius equal?

On June 18 2005 21:05 decafchicken wrote:
Found these math problems on google: enjoy
1. One day, a person went to horse racing area, Instead of counting the number of human and horses, he instead counted 74 heads and 196 legs. Yet he knew the number of humans and horses there. How did he do it, and how many humans and horses are there?

2. y = log x
If y = 10, then what is x?

3. 10*9*8*7*6*5*4*3*2*1 = 10!
Can this be true?! Why or why not?

4. If 1/2x +1/2(1/2x + 1/2(1/2x +1/2(1/2x + ... = y,
then x = ?

5. What place in this world can have their temperatures Fahrenheit and Celsius equal?

On June 18 2005 21:24 lightman wrote:
Here's another one:

the numbers are the lenght of the hexagon's sides. Find the radius of the circle

On June 18 2005 21:38 Sorrow_eyes wrote:


1 is easy
let x be men, y be horse number
x+y=74
2x+4y=196
solve

2. um... Y=LogX, 10^X=Y, 10^X=10, x is 1.

On June 18 2005 23:21 bangchucaibang wrote:
Maybe you're mistaken or im dumb
if y = log x, then x = 10^y, not the other way around, rite?
so x = 10^10?

On June 18 2005 23:21 bangchucaibang wrote:


Maybe you're mistaken or im dumb
if y = log x, then x = 10^y, not the other way around, rite?
so x = 10^10?

On June 19 2005 05:20 fearus wrote:
I dont know if this has been posted but my ex asked me this.

A= sinx
B = n
C = 6

Prove:

A/B = C

For n can be any real or imaginary number...

On June 18 2005 05:12 BigBalls wrote:
lordofdabu is right

zero is an integer, youre thinking of natural numbers malmis

On June 19 2005 13:57 RaGe-xG- wrote:


zero is a natural number

On June 19 2005 13:57 LordOfDabu wrote:
Must the two roots be distinct? If not, a = 4, b = 4, c = 1 will yield an equation with a double root of x = 0.5, for 4x^2 - 4x + 1 = (2x - 1)(2x - 1)

On June 19 2005 14:02 BigBalls wrote:


http://mathworld.wolfram.com/NaturalNumber.html


america doesnt include it (or at least every teacher and professor ive ever had doesnt), belgium might

On June 18 2005 20:57 lightman wrote:
(....)
Please let me know if you have any questions

On June 19 2005 13:10 LordOfDabu wrote:
Here's another problem that I like:

Four bugs (let's call them A, B, C, and D) occupy the corners of a square 10 inches on each side (A has the upper left, B has the upper right, C has the lower right, and D has the lower left). Simultaneously A crawls towards B, B towards C, C towards D, and D towards A. Assuming that all four crawl at the same constant rate, how far will each travel before they meet?

On June 19 2005 21:36 Sorrow_eyes wrote:


10 inchese ofcorse

and yeah, this one is classic, my dad told me this.

On June 19 2005 22:12 LordOfDabu wrote:
They will spiral towards the center of the square.

On June 19 2005 23:18 PaeZ wrote:
Im sorry i just cant help it, america IS A CONTINENT, that name comes from the explorer Americo Vespucio, who named it when they discovered it wasnt actually the Indias, and forgive me but USA is USA, America is all this part of the globe from Canada to Argentina, including the Caribbean and such
but nowdays most people refer to America = USA which is not true and is a mistake
P.S im not flamming the US and no i dont hate the US and such so dont misunderstand my post

On June 19 2005 14:02 BigBalls wrote:


http://mathworld.wolfram.com/NaturalNumber.html

america doesnt include it (or at least every teacher and professor ive ever had doesnt), belgium might

On June 19 2005 20:06 Bill307 wrote:
Question: On the planet Thassux, given two points {a, b} and {c, d}, what is the distance between them? Recall that on Earth, the distance between two points (a, b) and (c, d) is given by the square root of (a - b)^2 + (c - d)^2. You can leave your answer in terms of binomials like "(a - c)": it doesn't get any simpler if you expand them out.

On June 19 2005 20:06 Bill307 wrote:Also, for people interested in more of a challenge, try going in the opposite direction: as you have seen, the formula on the planet Thassux is not particularly elegant. However, on the planet Mophux, the distance from a point {x, y} to the origin {0, 0} is given by the square root of x^2 + xy + y^2. Assuming that, like Thassux and Earth, the first coordinate is how many units you must move "to the right", describe how the coordinate system may work on Mophux. If you want, you can also come up with other solutions that ignore this assumption, which would be cool .

On June 19 2005 23:27 loloko2 wrote:

I totally agree with him, you cant argue that guys, imsorry, hes right... ^_^

On June 20 2005 10:22 lightman wrote:


Don't be stupid and come up with all that Vespucio stuff.

On June 19 2005 23:24 Sorrow_eyes wrote:

pm me w/ ur Email addr for a UMS map on it :D

On June 20 2005 07:29 Catyoul wrote:
d² = (a-c)² + (b-d)² + sqrt(2) (a-c) (b-d)

On June 20 2005 07:29 Catyoul wrote:
With the assumption, second coordinate is 1/2 unit to the right, sqrt(3)/2 up

Without it, it's funnier, the family of solutions can be described like this :
Let a and b be 2 arbitrary numbers with the following relation : a - b = +/- pi/3 + 2 k pi (with k an integer)
Let u1 = cos a, u2 = sin a, v1 = cos b, v2 = sin b.
All coordinate systems on Mophux are of the form : u1 to the right and u2 to the top for the first coordinate, v1 to the right and v2 to the top for the second coordinate.

On June 20 2005 15:05 Cambium wrote:
Here is a good one:

In the diagram, two circles are tangent to each other at point B. A straight line is drawn through B cutting the two circles at A and C, as shown. Tangent lines are drawn to the circles at A and C. Prove that these two tangent lines are parallel.

On June 20 2005 13:03 lightman wrote:
Back to our interesting and nice thread. Here's a hard one:

Let n be a natural number. A cube with an arist of lenght n can be divided into 1996 cubes which arists are also natural numbers. Find the minor possible value of n.

On June 20 2005 15:05 Cambium wrote:
Here is a good one:



In the diagram, two circles are tangent to each other at point B. A straight line is drawn through B cutting the two circles at A and C, as shown. Tangent lines are drawn to the circles at A and C. Prove that these two tangent lines are parallel.