more or less than the others. What is the minimum number of weightings you need and how do you go about determining which is the false coin and wether it weighs more or less than the others?
2006-06-24 15:31:08 · 4 answers · asked by Pavi 2 in Science & Mathematics ➔ Mathematics
I think the time has come to make two comments about my question. First, chance or probabilities have nothing to do with the problem, Second, and this is a hint, the minimum of tries are three.
Pavi
2006-06-26 09:00:13 · update #1
Remember, you do not know in advance wether the false coin weighs MORE or LESS than the others.
2006-06-26 09:03:47 · update #2
4 weighings. First weigh 6 and 6. If one is heavy you take the heavy side. 3 and 3. Pick the heavy side. 1 and 1, if equal it is the third.
The 4th weighing is one of the coins may be light and when you weigh 3 vs 3 they might be the same and you would have to pick the other pile of 6.
2006-06-24 15:39:29 · answer #1 · answered by Nelson_DeVon 7 · 1⤊ 0⤋
The answer is 4.
1) Weigh 8 of the coins, 4 vs 4. (Let's label these coins 1-8) If these balance, then you can eliminate these 8 and you're down to the other 4 (coins 9-12); otherwise you're down to 8 (1-8), but note which 4 are heavier and which 4 are lighter.
If you were fortunate in step 1 and are down to 4 coins (9-12), figuring out which coin it is and whether or not it is heavier or lighter than the others can easily be done in three more steps.
4-2) Compare coins 9 & 10. If they're the same it's down to 11 & 12. If they're different it's 9 & 10.
4-3) Weigh one of the two remaining coins against a known good coin. Now you know which coin it is.
4-4) If you don't know heavier/lighter, weigh the coin against a known coin.
Now let's revisit the scenario from step 1 where 8 coins were remaining.
8-2) Take the heavier coins (label them 1-4) and weigh them. If they balance then, one of the lighter coins (coins 5-8) is the different one. If they don't balance, it's one of the heavier coins. At this point you're down to 4 coins, and you know if the coin is heavier or lighter.
8-3) Weigh 2 of the remaining coins - 1 vs. 1. At this point you can get down to 2 coins in the worst case.
8-4) Weigh these 2 coins and since you know heavier/lighter from step 8-2, select the heavier/lighter one.
Maximum tries necessary should never be more than 4.
2006-06-24 23:23:04 · answer #2 · answered by Mark M 1 · 0⤊ 0⤋
Minimum number of weighings is four if you assume that chance always goes against you. Start by assuming the odd coin is heavier than the rest. If you guess right, then it will only take 3 weighings. Otherwise it takes 4.
Case 1 - Odd coin is heavier:
1st weighing, put all coins on the scale. 6 on each side. They won't weigh the same.
2nd weighing, take the 6 coins from the heavier stack, and put 3 on each side of the scale. They won't weigh the same. This shows the odd coin is heavier than the rest.
3rd weighing, take 2 of the 3 coins from the heavier stack, and put one on each side of the scale. The heavier of the two is the odd coin. If they weigh the same, the odd coin is the one that wasn't weighed.
Case 2 - Odd coin is lighter:
1st weighing, put all coins on the scale. 6 on each side. They won't weigh the same.
2nd weighing, take the 6 coins from the heavier stack, and put 3 on each side of the scale. They will weigh the same. This shows the odd coin is lighter than the rest.
3rd weighing, take the 6 coins from the lighter stack, and put 3 on each side of the scale. They won't weigh the same.
4th weighing, take 2 of the 3 coins from the lighter stack, and put one on each side of the scale. The lighter of the two is the odd coin. If they weigh the same, the odd coin is the one that wasn't weighed.
The method above will always find the odd coin in 3 or 4 weighings. You could use a different method where, if you get lucky, you could prove which is the odd coin in 2 weighings (2 is the absolute minimum since you don't know if the odd coin is lighter or heavier), but it will take 5 weighings to prove which is the odd coin if you assume worst case luck. I figure the method I showed is what you're looking for.
2006-06-25 04:43:35 · answer #3 · answered by tom_2727 5 · 0⤊ 0⤋
you have 1 to 12
Divide the 12 in half
1 to 6 : 7 to 12
Whichever side is heavier, divide that in half
1 to 3 : 4 to 6
Whichever side is heavier, divide that in half
1 : 2 with 3 left out
if 1 is heavier, then 1 is the answer
if 2 is heavier, then 2 is the answer
if 1 and 2 are evenly balanced, then 3 is the answer
ANS : 3 times
2006-06-24 22:45:20 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋