You have 12 coins and a balance. One of the coins is either heavier or lighter than the others. Show the logical steps to come up with the solution. If you can figure this out, you are a damn genius!
A Brain Teaser NO ONE will be able to get
- Thread starter HereThereandEverywhere
- Start date
Weigh each coin one at a time. Whichever is lightest is the light one.
I think maybe you an integral part of the problem: Restrictions.
I think maybe you an integral part of the problem: Restrictions.
I have seen something like this before and IIRC then you had to find which was the odd coin by only using the balance three times.
I said a balance, not a scale. A balance is something that will tell you which side is heavier than the other side.
why is this even a problem?
even with 3 time restriction
6 coins vs 6 coins.
than one side is gonna be lighter
so 3 vs3
than left with 3 coins
1v1. either 1 is lighter, or =, than its the 3rd.
even with 3 time restriction
6 coins vs 6 coins.
than one side is gonna be lighter
so 3 vs3
than left with 3 coins
1v1. either 1 is lighter, or =, than its the 3rd.
According to this version you can only use the balance three times.
It has a solution too, for those who don't want to figure it out themselves.
It has a solution too, for those who don't want to figure it out themselves.
Slacker
Diamond Member
- Oct 9, 1999
- 8,623
- 33
- 91
need more rules, to take it exactly as you wrote it there are several solutions
what are the weights of the coins? if that is up to me to decide then I will say that 11 of them are 100 grams each and one of them is 99.99999 grams , then what is the sensitivity of the scale? I can add more "9's" to the number of the lighter coin until I reach the balance sensitivity point
what are the weights of the coins? if that is up to me to decide then I will say that 11 of them are 100 grams each and one of them is 99.99999 grams , then what is the sensitivity of the scale? I can add more "9's" to the number of the lighter coin until I reach the balance sensitivity point
littleprince, the method you suggested would be great if you knew that the counterfeit coin was definitely lighter than the other 11.
It can be lighter or heavier, so you wouldn't know which of the groups of six contained the counterfeit on the first weigh.
It can be lighter or heavier, so you wouldn't know which of the groups of six contained the counterfeit on the first weigh.
IJump
Diamond Member
- Feb 12, 2001
- 4,640
- 11
- 76
<< why is this even a problem?
even with 3 time restriction
6 coins vs 6 coins.
than one side is gonna be lighter
so 3 vs3
than left with 3 coins
1v1. either 1 is lighter, or =, than its the 3rd. >>
That is the correct answer for using the scale the fewest number of times...............
littleprince, it's not quite as easy as that since he did say that one of the coins is lighter or heavier than the others.
[edit]: oops, didn't see Haircut's message... but it applies to what IJump said too.
[edit]: oops, didn't see Haircut's message...
but it applies to what IJump said too.Too bad the thread author couldn't even get the teaser's rules right. I see someone posted the correct answer to this 'unsolvable brain teaser' in under 10 minutes. sigh
Lithium381
Lifer
- May 12, 2001
- 12,455
- 7
- 81
heh, i geuss i'm a geniuos, i figured it out! wow that was easy....even with that three times rule.......
<< what are the weights of the coins? if that is up to me to decide then I will say that 11 of them are 100 grams each and one of them is 99.99999 grams , then what is the sensitivity of the scale? I can add more "9's" to the number of the lighter coin until I reach the balance sensitivity point >>
Give it a rest, eh?
1. Place the balance on a flat surface with empty bowls.
2. Take 2 coins and place 1 coin in each bowl.
3. Repeat step 2 until the balance is not in balance anymore.
4. IF balance NOT in balance THEN 1 of the 2 last coins is different,
2. Take 2 coins and place 1 coin in each bowl.
3. Repeat step 2 until the balance is not in balance anymore.
4. IF balance NOT in balance THEN 1 of the 2 last coins is different,
The version I'm familiar with has 27 coins, and three shots with a balance to find the one coin that is the lighter/heavier one. Can anyone figure this one out? 12 coins is too easy.
Divide in 3 parts.
Put 1st pair of 9 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 9 coins.
If not in balance, 1 of the piles on the scale contains the odd coin.
Take the pile of 9 coins containing the odd coin and divide in 3 piles of 3 coins.
Put 1st pair of 3 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 3 coins.
If not in balance, 1 of the piles on the scale contains the odd coin.
Take a pair of coins from the pile of 3 coins containg the odd coin and put 1 in each bowl of the balance.
If not in balance the 3'rd coin is odd.
If in balance, 1 of the coins on the balance is odd.
3 shots with the balance, right?
Put 1st pair of 9 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 9 coins.
If not in balance, 1 of the piles on the scale contains the odd coin.
Take the pile of 9 coins containing the odd coin and divide in 3 piles of 3 coins.
Put 1st pair of 3 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 3 coins.
If not in balance, 1 of the piles on the scale contains the odd coin.
Take a pair of coins from the pile of 3 coins containg the odd coin and put 1 in each bowl of the balance.
If not in balance the 3'rd coin is odd.
If in balance, 1 of the coins on the balance is odd.
3 shots with the balance, right?
Ugh. I actually took the time to figuire this out not too long ago. However typing out the solution takes almost as long as actually figuiring it out. If you want the answer I am sure you can google it.
<< Divide in 3 parts.
Put 1st pair of 9 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 9 coins.
If not in balance, 1 of the piles on the scale contains the odd coin. >>
hmmm... yes, it sounds good so far. But I get stuck when the scale is not balanced. Which set of 9 coins do I proceed with? I don't know if the coin is supposed to be lighter or heavier.
<< Take the pile of 9 coins containing the odd coin and divide in 3 piles of 3 coins.
Put 1st pair of 3 coins on the balance.
If it is in balance the odd coin is in the 3'rd pile of 3 coins.
If not in balance, 1 of the piles on the scale contains the odd coin. >>
similar problem as above.
<< Take a pair of coins from the pile of 3 coins containg the odd coin and put 1 in each bowl of the balance.
If not in balance the 3'rd coin is odd.
If in balance, 1 of the coins on the balance is odd.
3 shots with the balance, right?
>>I can do it with 30 million coins and 1 different one:
Go to the bank, and trade it in for banknotes, the different coin will be rejected by the counting machine.
But this problem came along a few months ago, and as I said then, it is not possible to do in 3 times, unless you're extremely lucky (and then it's possible in 2 times too )
Go to the bank, and trade it in for banknotes, the different coin will be rejected by the counting machine.
But this problem came along a few months ago, and as I said then, it is not possible to do in 3 times, unless you're extremely lucky (and then it's possible in 2 times too
)
UberNeuman
Lifer
- Nov 4, 1999
- 16,937
- 3,087
- 126
TRENDING THREADS
-
Discussion Zen 5 Speculation (EPYC Turin and Strix Point/Granite Ridge - Ryzen 9000)- Started by DisEnchantment
- Replies: 25K
-
Discussion Intel Meteor, Arrow, Lunar & Panther Lakes + WCL Discussion Threads
- Started by Tigerick
- Replies: 24K
-
Discussion Intel current and future Lakes & Rapids thread- Started by TheF34RChannel
- Replies: 23K
-
-
Facebook
Twitter