Coin Conundrum
You have a pile of 24 coins. Twenty-three of these coins have the same weight, and one is heavier. Your task is to determine which coin is heavier and do so in the minimum number of weighings. You are given a beam balance (scale), which will compare the weight of any two sets of coins out of the total set of 24 coins. How many weighings are required to identify the heavier coin?
-Rx
-Rx
Labels: logic, SharedPuzzle
posted by Mr Fahrenheit at
12:22 AM
12 Comments:
at most 4 times.
24/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1 or 2
3 times
1. divide to 3 piles of 8. Weigh 2 piles. If one is heavier, continue, if equal, the one left on the side has the heavy coin
2. divide pile of 8 to 3, 3, & 2. weigh both piles of 3, if one is heavier, continue, if equal, the pile of 2 has the heavy coin
3. If pile of 3 is heavier, weigh 2 of the coins - either one of them is the heavy coin, or its the one on the side. If pile of 2 coins from #2 is left, just weigh both.
dang.. arbel had my answer..
3 weighings..
1. put 8 coins on each side.. if even, discard these 16, if not, keep heaver and discard rest (8 left)
2. put 3 coins on each side.. if even, discard these 6, if not, keep heaver and discard the rest (3 or 2 left)
3. put 1 coin on each side.. if even, discard both and take left over, if not, take the heavier.
1
guess and get it right :)
but arbel is right
manjinder
weigh six, six, six, and six. Through away three of the four sets of sixes that weight the same and weigh two, two, and two for the remaining 6.
Again, throw away two of the three sets of two that weighs the same and you are left with 2. Weigh these to find the heaviest.
Answer = 3 tries.
Zero times you can look at the coins to see which is bigger. Bigger = Heavyer
What if one is gold and the others are bronze, they are all the same size, and are all painted black?
3 weighnigs...
24
|
8 8 8-- 1st weighing
|
3 3 3--- 2nd weighing
|
1 1 1 ---- 3rd weighing
anonymous3: bigger aint heavier! thats like saying a beach ball is heavier than a bowling ball! more mass = more weight!
i say four trips
1.split to 12 (2 piles)
2.take heavier pile split into two pile of 6 and weigh
3.take heavier then split into two piles of 3 and weigh
4. take heaviest pile and split into three sets of 1 and weigh two, wichever is heaviest is the heaviest!
o<-< stickman
I knew this from some time but had forgotten how to get it right (there's even a "Columbo" episode with this problem given to columbo by a suspect, and solved by ... his wife as usual).
The correct answer given is the 8 8 8 start, then 3 3 2 (weighting 3 3 together) then either 1 1 or 1 1 1 (depending if one of the 3 set was heavier or not)
whatever on the third weighting you can determine which is the heaviest coin.
Dichotomia (beginning with 12 12 set then 6 6 then 3 3 then 1 1 1) is a good general answer but not the optimal one in this case.
Then I wondered if trying a 6 6 6 6 beginning could be good... the answer is obviously no. Because you can't beat the 8 8 8 and you can be in the 12 12 disposition .
Imagine one of the 6 is heavier that's the best disposition whatever with 6 coin you have at least 2 weighting before finding the good coin (either doing 3 3 and 1 1 1 or doing 2 2 2 and 1 1.. can't do better you can only compare 2 sets at once). Then if in the first weighting none is heavier you're left with 2 6 sets as in the 12 12 disposition (meaning you have a litle chance to do like the 12 12 disposition and a little chance to do like the 8 8 8( actually it's a 1/2 probability so I guess you could call it a 3.5 weighting configuration ;).
Why 8 8 8 better? It seems that eliminating 16 coins in the first weighting with the 8 8 8 configuration is really great I guess ;).
Yeah I know it's not really all clear as an explaination and it certainly is not a good mathematical proof but well I tried to explain how I understood it.
-Funky MnX-
Arbel is correct!
-Rx
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