Capt. Orygun
Win the Day
CAL wins
Brain teaser
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Originally posted by SiR Kid+Jun 10 2004, 04:19 AM-->
There is no way to tell from the information given.
BUT, the fastest way to find out which, and if heavier or lighter is listed here.
Weight 4 and 4. Which side is heavier?
Remove 1 of each side.
Weigh 3 and 3. Which side is heavier? If same, then the coins you removed from the previous step were different from each other. Weigh one against one of the remaining 6 to find the oddball.
If the 3 and 3 are still different, remove one from each side and weigh again.
Repeat til you find the oddball.
that takes way more than 3 uses of the scale unless you get lucky. there is a real solution... that is not it.
You can't tell if the odd one is lighter or heavier with the information you gave. PERIOD.
The way to get the odd coin in three uses is measure 4 and 4, take one side, split them and weigh 2 and 2.
Take the one side and weigh 1 and 1.
Done. You found the odd coin. BUT this is ONLY if you know wether the coin is ligher or heavier to begin with.
The way to get the odd coin in three uses is measure 4 and 4, take one side, split them and weigh 2 and 2.
Take the one side and weigh 1 and 1.
Done. You found the odd coin. BUT this is ONLY if you know wether the coin is ligher or heavier to begin with.
Originally posted by Silverchild79@Jun 10 2004, 05:46 AM
CAL wins
I like the torus solution better, but I wouldn't call it a 2D solution. It's a 2D surface, but still uses 3D positioning. I guess it all depends on your coordinate system.
Capt. Orygun
Win the Day
no you:
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXT
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXT
Originally posted by Silverchild79@Jun 10 2004, 12:24 PM
no you:
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXT
nope. if you put 4 on each side, you don't know if the coin is lighter or heavier so how do you know what side to discard?
THERE IS AN ANSWER. i promise you.
Frankie P.
Senior Member
How can one weigh more, and then the coins be balanced out? If you spilt up the 8 coins by putting 4 on each side, one will side will weigh more. You remove all the coins from the lighter side, and split the 4 heavier into 2 on each side. Then you discard the lighter size, weigh the 2 heavy ones seperately and you find the heavier one. But if there is a heavier coin, but it weighs the same I think it is impossible because it is a trick question.
Originally posted by ndogg@Jun 10 2004, 12:05 AM
just bring the line through the house not under it. then its 2-d. its a stupid trick question any way you look at it. if you look at it with graph theory (wich i have done) there is no solution. unless you use a special 2-d surface called a torus:
Cheater!! He got that picture from the same site I used to cheat....
..I mean, uh, verify my results.
Solution to the coin problem:
Take four of the coins and put them on a scale: two against two
12 – 34 (unused coins: 5678)
If they balance, you know you have four “honest†coins.
For this example, let’s take the longest path to a solution. So, let’s assume that coins 1234 balanced, therefore, we know our alien coin is either 5,6,7 or 8.
Balance three of the “good†coins (123) against three uncertain coins (567)
123 – 567
If they balance, we know coin 8 is the alien, and weighing it against a good coin will tell us if it’s heavier or lighter.
If they don’t balance, we know that 5, 6 or 7 is the bad coin, AND we know whether the alien coin is lighter or heavier by which direction the pile goes relative to the three good coins. (Let’s assume the 567 is lighter, meaning that the alien coin is lighter)
To determine which of 5, 6 or 7 is bad, simply weigh 5 and 6 against each other. If they don’t balance, then the lighter one is bad. If they balance, then we know 8 is bad – and lighter.
Take four of the coins and put them on a scale: two against two
12 – 34 (unused coins: 5678)
If they balance, you know you have four “honest†coins.
For this example, let’s take the longest path to a solution. So, let’s assume that coins 1234 balanced, therefore, we know our alien coin is either 5,6,7 or 8.
Balance three of the “good†coins (123) against three uncertain coins (567)
123 – 567
If they balance, we know coin 8 is the alien, and weighing it against a good coin will tell us if it’s heavier or lighter.
If they don’t balance, we know that 5, 6 or 7 is the bad coin, AND we know whether the alien coin is lighter or heavier by which direction the pile goes relative to the three good coins. (Let’s assume the 567 is lighter, meaning that the alien coin is lighter)
To determine which of 5, 6 or 7 is bad, simply weigh 5 and 6 against each other. If they don’t balance, then the lighter one is bad. If they balance, then we know 8 is bad – and lighter.
Capt. Orygun
Win the Day
Originally posted by ndogg+Jun 10 2004, 12:45 PM-->@Jun 10 2004, 12:24 PM
no you:
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXT
nope. if you put 4 on each side, you don't know if the coin is lighter or heavier so how do you know what side to discard?
THERE IS AN ANSWER. i promise you.
it's impossible to do in 3 uses of the scale unless you get lucky or you can tell wich is diff with your hands
Originally posted by Silverchild79+Jun 10 2004, 03:52 PM-->@Jun 10 2004, 12:24 PM
no you:
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXTÂ
nope. if you put 4 on each side, you don't know if the coin is lighter or heavier so how do you know what side to discard?
THERE IS AN ANSWER. i promise you.
it's impossible to do in 3 uses of the scale unless you get lucky or you can tell wich is diff with your hands
very do-able.
Originally posted by Silverchild79+Jun 10 2004, 03:52 PM-->@Jun 10 2004, 12:24 PM
no you:
1. wieght the coins four a side, one side will be heavier, discard the light side (1 scale use)
2. Split the remianing four coins into stacks of two and weight them, discard the lighter stack. You should have two coins left (scale use two)
3. Of the remianing two coins determine which is heavier with your last use of the scale (3)
NEXTÂ
nope. if you put 4 on each side, you don't know if the coin is lighter or heavier so how do you know what side to discard?
THERE IS AN ANSWER. i promise you.
it's impossible to do in 3 uses of the scale unless you get lucky or you can tell wich is diff with your hands
d00d, look like 4 post up from here. the solution is posted. good job number9!
Originally posted by Airjockie@Jun 10 2004, 04:45 PM
OK.....
Now tell me how a bumble bee can fly....
It is litterally impossible...the wings are to small, and the body is too big and heavy...but yet it can fly....but how
now no googleing....
I'm more impressed that 747s get off the ground. Now *that's* amazing!
I was visiting a friend recently who lives just west of O'Hare. I was watching those big birds flying overhead on their way to a landing as gentle as can be.
I think I recall reading the bumble bee's secret somewhere. But I don't remember it now.
