|
On April 10 2011 10:36 Everhate wrote:Show nested quote +On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
|
[QUOTE]On April 10 2011 10:39 Darkren wrote:
[QUOTE]On April 10 2011 09:41 Frigo wrote:
[QUOTE]On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
[/quote]
You sure this is possible?
[quote]
[/QUOTE]
Yes it's possible ill give a hint.
+ Show Spoiler +U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it[/QUOTE]
If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs...
|
+ Show Spoiler +1:
Pour 5 liters into 5 liter jug.
Fill up the 3 liter jug using water from the 5 liter jug.
You have 2 liters of water in the 5 liter jug now.
Pour out the 3 liter jug.
Pour the 2 liters of water in the 5 liter jug into the 3 liter jug.
Fill the 5 liter jug again.
Fill the 3 liter jug with water from the 5 liter jug.
The 5 liter jug has 4 liters in it.
|
|
|
|
|
On April 10 2011 05:08 shadowy wrote:Show nested quote +On April 10 2011 01:54 Kazius wrote:Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this): 12. + Show Spoiler +You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door . Wrong logic  + Show Spoiler +
Once of the doors is open, the chances now are 50/50. It doesn't matter if you switch or not - there is no better or worse chance to get it right.
Edit: To explain it better: There are now 2 closed doors, 1 car and 1 goat. It's 1/2 chance to get the car...or a goat.
the first reasoning is correct.
|
[QUOTE]On April 10 2011 10:44 ixi.genocide wrote:
[QUOTE]On April 10 2011 10:39 Darkren wrote:
[QUOTE]On April 10 2011 09:41 Frigo wrote:
[QUOTE]On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
[/quote]
You sure this is possible?
[quote]
[/QUOTE]
Yes it's possible ill give a hint.
+ Show Spoiler +U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it[/QUOTE]
If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs...
[/QUOTE]
No it's possible, i gave u the first part of the solution, this isn't a 2 min riddle u have to think on it.
|
are u sure dude? have ur calculator a go.
|
On April 10 2011 10:59 Darkren wrote:Show nested quote +On April 10 2011 10:44 ixi.genocide wrote:On April 10 2011 10:39 Darkren wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? Yes it's possible ill give a hint. + Show Spoiler +U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4 If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs...
No it's possible, i gave u the first part of the solution, this isn't a 2 min riddle u have to think on it.[/QUOTE]
|
Fenrax
United States5018 Posts
This thread becomes a mess. Plz just post your solution in Spoiler Tags in the same post as the question.
|
On April 10 2011 11:00 sicajung wrote:are u sure dude? have ur calculator a go.
Yes i am sure.
+ Show Spoiler +Ur getting caught in the problem, multiplication and division alwais take the priority, the equation can be rewroten (48/2)(9+3) which equals 24 X 12 = 288, ur calculator can't see these errors
|
On April 10 2011 10:43 Darkren wrote:Show nested quote +On April 10 2011 10:36 Everhate wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
+ Show Spoiler +
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B from the opposite side is heavy. If they do not balance, the lighter one is the incorrect weight.
|
|
|
23.+ Show Spoiler +12 coins, 1 of which is either heavy or light
EEEEEEEEEEEE (24 possibilities), measure 4v4
they balance, the 4 others are either heavy or light
___EEEE (8 possibilities), measure 3 versus 3 legit
___ they balance, the one you didn't measure is either heavy or light
______E (2 possibilities), measure it versus 1 legit
___ they don't, you have 3 heavy or 3 light
______hhh or lll (3 possibilities), compare 2
______ they balance, the third is the fake
______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light
___hhhhllll (8 possibilities), measure hhl versus hhl
___ they balance, one of the 2 lights you didn't measure is fake
______ll (2 possibilities), measure versus each other
___ they don't, one of the 2 heavies was heavy or the opposite light was light
______hhl (3 possibilities), measure the heavies versus each other
______ they balance, the light was light
______ they don't, the heavy one is heavy
With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
|
On April 10 2011 11:35 Everhate wrote:Show nested quote +On April 10 2011 10:43 Darkren wrote:On April 10 2011 10:36 Everhate wrote:On April 10 2011 09:41 Frigo wrote:On April 10 2011 08:43 Darkren wrote:
Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible? I believe this approach will work...unless someone can point out the flaw: + Show Spoiler +
Place 3 coins on each side, set the rest aside
-> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight.
Of the 6 unknown, place 2 on each side of the scale
-> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight.
Place one of the remaining coins on one side, with one of the known coins on the other.
-> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p + Show Spoiler +
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
Ur close but not quite there
+ Show Spoiler +On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above
|
On April 10 2011 11:46 gyth wrote:23. + Show Spoiler +12 coins, 1 of which is either heavy or light
EEEEEEEEEEEE (24 possibilities), measure 4v4
they balance, the 4 others are either heavy or light
___EEEE (8 possibilities), measure 3 versus 3 legit
___ they balance, the one you didn't measure is either heavy or light
______E (2 possibilities), measure it versus 1 legit
___ they don't, you have 3 heavy or 3 light
______hhh or lll (3 possibilities), compare 2
______ they balance, the third is the fake
______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light
___hhhhllll (8 possibilities), measure hhl versus hhl
___ they balance, one of the 2 lights you didn't measure is fake
______ll (2 possibilities), measure versus each other
___ they don't, one of the 2 heavies was heavy or the opposite light was light
______hhl (3 possibilities), measure the heavies versus each other
______ they balance, the light was light
______ they don't, the heavy one is heavy With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
Got it
|
Fenrax
United States5018 Posts
On April 10 2011 11:46 gyth wrote:23. + Show Spoiler +12 coins, 1 of which is either heavy or light
EEEEEEEEEEEE (24 possibilities), measure 4v4
they balance, the 4 others are either heavy or light
___EEEE (8 possibilities), measure 3 versus 3 legit
___ they balance, the one you didn't measure is either heavy or light
______E (2 possibilities), measure it versus 1 legit
___ they don't, you have 3 heavy or 3 light
______hhh or lll (3 possibilities), compare 2
______ they balance, the third is the fake
______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light
___hhhhllll (8 possibilities), measure hhl versus hhl
___ they balance, one of the 2 lights you didn't measure is fake
______ll (2 possibilities), measure versus each other
___ they don't, one of the 2 heavies was heavy or the opposite light was light
______hhl (3 possibilities), measure the heavies versus each other
______ they balance, the light was light
______ they don't, the heavy one is heavy With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
I don't think it is even possible to find a solution WITHOUT telling if the counterfeit is heavy or light.
|
On April 10 2011 00:05 stepover12 wrote:19.(pirates, arrrr) + Show Spoiler + On April 10 2011 06:32 tomnov wrote:10 pirates found a loot of 100 gold pieces, and decided to split it the following way:
the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on.
the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die
* a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
I'll add more when I can.
Ive had my math teacher ask me this riddle but with the twist how much has the first pirate offer to not be killed and make the most money. Apparently he can make in the range of 30$-40$.
Ive not yet solved it any imput would be appreciated
|
On April 10 2011 11:44 sicajung wrote:Show nested quote +On April 10 2011 11:26 Darkren wrote:On April 10 2011 11:00 sicajung wrote:On April 10 2011 10:57 Darkren wrote:On April 10 2011 10:54 sicajung wrote:
48/2(9+3)
solve this. + Show Spoiler + are u sure dude? have ur calculator a go. Yes i am sure. + Show Spoiler +Ur getting caught in the problem, multiplication and division alwais take the priority, the equation can be rewroten (48/2)(9+3) which equals 24 X 12 = 288, ur calculator can't see these errors just wanna open up ur mind :D http://mybroadband.co.za/vb/showthread.php/325471-48÷2-(9-3)-argument
this stuff is purely notation ambiguity, no real substance man.
|
On April 10 2011 12:06 Fenrax wrote:
I don't think it is even possible to find a solution WITHOUT telling if the counterfeit is heavy or light.
In one of the solutions, if everything always balanced, then you'd know the last coin was fake, but not whether it was heavy or light.
|
|
|
|
|
|
EPT
