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Light one string at one end and the other string at both ends at 0:00. When the second string burns out it will be 0:30; light the first string from the other end. When that one's completely gone too, it's 0:45, you Amish pyro. Burning precious string-clocks? Shame on you!
:D hah! that's right!Quote:
Originally posted by solarflare
Light one string at one end and the other string at both ends at 0:00. When the second string burns out it will be 0:30; light the first string from the other end. When that one's completely gone too, it's 0:45, you Amish pyro. Burning precious string-clocks? Shame on you!
So do you have a questoin?
Do I? (Hint: there are only two possible answers and 'yes' is wrong)Quote:
Originally posted by SeventhStar
:D hah! that's right!
So do you have a questoin?
I'll try to revive this! Again ;)
A warden of a prison comes back to the prison from a party very drunk. In his state, he accidentally turns the key to open all 100 of the jail cells.He realizes what he's done, so he goes back and turns the key on every other cell (thus, now cell 1 is open, cell 2 is closed, cell 3 is open, etc.). He realizes that he hasn't closed all of them, so he goes back and turns thekey on every third cell (thus, now cell 3 is closed, and cell 6, which heclosed when he turned every other key, is now open). Still drunk, he goesback and turns the key on every 4th, then every 5th, 6th, 7th, etc. until heturns the key on every 100th cell (thus, only on cell #100). After all this,which cells remain open?
The first one.
nope! That's not correct :) sorryQuote:
Originally posted by Elrond
The first one.
Every cell that whose # happens to be a perfect sqare. In other words:
The 1st cell, 4th cell, 9th cell, 16th cell, 25th cell, 36th cell, 49th cell, 64th cell, 81st cell, and 100th cell are all open.
- Kevin
P.S. The prisoners must have been twice as drunk as the prison guard if they didn't wake up from all the noise and escape when they had the chance. :D
Uhm.. a keik:) :p
Just a sec for a joke...:)Sorry
yep That's right :DQuote:
Originally posted by KevinHall
Every cell that whose # happens to be a perfect sqare. In other words:
The 1st cell, 4th cell, 9th cell, 16th cell, 25th cell, 36th cell, 49th cell, 64th cell, 81st cell, and 100th cell are all open.
- Kevin
Heh :D That's right tooQuote:
Originally posted by KevinHall
P.S. The prisoners must have been twice as drunk as the prison guard if they didn't wake up from all the noise and escape when they had the chance. :D
I forfeit my turn at authoring a question. Anyone got a new question?
I am sorry I wasso wrong...:)
Ok I hawve a new one :D
A monk needed to go from his monastery to another, located much higher in the mountains. So off he went. Sometimes he went fast, sometimes slow, but finally he got there. He finished his job there and started going back to his monastery. He reached it a lot faster because he was descending instead of climbing. As a whole he went really uneven in both directions. So question is: If the monk's monastery is called A and the other one is B is there a point (C) that is positioned in time so that AC (in time) is equal BC (in time) in the monk's walk.
Prove your suggestions.
There is.
On his way back, obviously at his start, the lenght of time between B and his position (P) is zeor, that is smaller than the time between A and P: t(BP) < t(AP).
The more he goes down, the higher t(BP) is (as time pass), and the smaller t(AP) is (as he is going closer from his start from A).
When he reaches A, t(AP) is 0, while t(BP) his higher than that: t(BP) > t(AP).
There must be a point somewhere where t(BP) goes from less than to more than t(AP): t(BP) == t(AP). This point is C.
can you tell how to find that pointQuote:
Originally posted by Elrond
There is.
On his way back, obviously at his start, the lenght of time between B and his position (P) is zeor, that is smaller than the time between A and P: t(BP) < t(AP).
The more he goes down, the higher t(BP) is (as time pass), and the smaller t(AP) is (as he is going closer from his start from A).
When he reaches A, t(AP) is 0, while t(BP) his higher than that: t(BP) > t(AP).
There must be a point somewhere where t(BP) goes from less than to more than t(AP): t(BP) == t(AP). This point is C.
400Km