The counterfeit coins?
You have 10 stacks of coins and each stack contains 10 coins. One complete stack of coins is counterfeit. You know the weight of each genuine coin and you know that the weight of each counterfeit coin is 1 gram heavier or lighter than a genuine coin.
Using a pointer scale ie one that gives an actual reading (not a balance), what is the minimum amount of weighings you would need to do to determine which stack is counterfeit?
You can split the stacks up if you want when weighing them.
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It can be done in one weighing.
Suppose each genuine coin weighs x grams
Take a sample of 1 coin from pile 1, 2 from Pile 2 etc: 55 coins in all
Compare weight of sample with 55 x
It will be heavier or lighter than 55 x by y grams, and y is the number of the pile of dud coins
1. Label the stacks 1-10
2. Take 1 coin from stack 1, 2 from stack 2, etc.
3. Weigh.
4. Subtract 55 from the total weight, you will have the number of the counterfeit stack.
i think the minimum amount of weighings is 6. you split the 10 stacks up into 3,3, and 4. you weigh the first 3. weigh the second 3. weight the group of 4. the longest possible outcome would be the group of 4. so you split that into two again. weigh one stack of two. figuring out which of the "2" the counterfeit stack is in. weigh each of the two and the one with the larger reading is the counterfeit one.
If you can breach the sacks then the answer has already been given – otherwise the answer is 5…
Select any five sacks and weigh them altogether [1]
Now weigh the other five [2]
the heavier of the two contains the counterfeit coins, so
take any two from the heavier five, weigh them together [3]
take another two from the remaining three – weigh them [4]
If weighings 3 and 4 are identical then the last remaining sack from the heaviest five contains counterfeit coins.
Otherwise, one final weighing – one of the two sacks from the heavier of weighings 3 and 4. If this is more than half of either weighing 3 or 4, then it has the bogus coins – else it’s the other one!