# Tag Info

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The solution is simple:

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It seems to me that there's a simpler solution than the one accepted above. Step 1: Step 2: Step 3: The point here is that

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First weigh two of your objects against the other two. Whichever pair is heavier must contain the 11-oz object, since even $11+3>5+8$. Now you have two objects of which you know one weighs 11 oz. Weigh them against each other to find out which one it is. Weigh the 11-oz object against two of the remaining three. If the scales balance exactly, those two ...

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You can number stacks from 1 to 10. P.S. You can do the same even if you have 11 stacks of 10: just number them from 0 to 10.

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In the worst-case scenario, it requires to locate the radioactive rods. Several answers already describe strategies for locating the radioactive rods. I will give another. Testing strategy: Start by testing two rods. If neither of these rods is radioactive, use the five remaining tests on five of the six remaining rods, one at a time. If two of these rods ...

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I see, it took me too long to fininsh my drawing, but let me present it as additional material to sousben's answer:

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You can: Note: Corrected after comment by Jaap Scherphuis.

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As many have already stated, the best you can do is Suppose the genuine coins have weight 0, and the other coins have weights of distinct powers of two. Then any try on the old balance will always just tip the scale to the side of the heaviest fake coin, since it weighs more than all the other coins on the scale put together. Now suppose the scale ...

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Yes. Solution:

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I can manage it in 8 weighings. I started with mdc32's answer and expanded on it. I was able to weigh 6 coins in 4 weighings. I suspect 12 can be done in 7 weighings, but I haven't figured out a way to do that yet. Start with a group of coins $\{c_1,c_2,c_3,c_4,c_5,c_6\}$. Weigh 1: $c_1 + c_2$. If the weight is either 20 or 40, follow mdc32's answer to ...

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Divide them into 3 groups of 4 people. Put any two groups on each side of the see-saw. (First Use) Condition 1 If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4. In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. (Second Use) Condition 1....

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Some of the existing answers to this ancient question are excellent, but there's one famous answer that I think deserves mention here. It comes from an article in Eureka, the annual magazine of the University of Cambridge's student mathematical society, written by C A B Smith under the pseudonym of "Blanche Descartes". It has two very nice features. The ...

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The greatest X for which you can find the stack with the fake coins in 3 weighings is: Unfortunately, the strategy isn't as easy to describe as the one in my previous answer (you can read it in the edit history if it helps understanding this one). Here it is: An example on how to interpret the weighing results: To brute-force find the selections (a list ...

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Since the weighing scale can only be used once, I feel free to disassemble and modify it. Now, tie a ball from each tumbler to the 4 vertices of the weighing-scale, as shown in this picture: Now, pull up the structure and see how it tilts: the highest ball from the floor is the lightest one!

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How many weighings you need: Call the four coins A,B,C and D, and the true gold coin G. You start by weighing Furthermore, it can't be done in just one weighing. There are nine possible situations (each coin could be heavy or light, or all could be same), and a weighing has only three outcomes. By the pigeonhole principle, there will be some outcome which ...

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I have found a set where the weight of the heaviest box is Here are the weights of the boxes General Strategy Minimality confirmed by Oray using computation.

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This is my first answer on the puzzling stack exchange and I'm really not used to explaining things like this so it's probably going to be rather convoluted. I'll probably come back and edit it later when I can figure out how to make this clearer. Solution for a maximum of 4 weightings: First divide the coins into 4 equally sized piles. There are now 3 ...

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There is a solution requiring a maximum of $9$ weighings. The strategy rests on a solution for $4$ coins $\{c_1,c_2,c_3,c_4\}$ in $3$ weighings - then performing the same operation on the remaining sets of $4$. With 4 coins, weigh coins $c_1$ and $c_2$. If the amount is 20 or 40, then the coins weigh 10 or 20, respectively, and we can just weigh the ...

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Fredo should weigh the single coin against nothing, then all 2016 coins together against nothing. If the coin is real, then 2016 times its weight will be within 99 grams of the total weight. Otherwise, it will be at least 1917 grams off.

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You only need Assume the genuine coins weigh $x$ grammes. Credit to Hexomino and Jaap for the corrections!

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Using all three weights in each trial is an interesting requirement, I haven't seen such a requirement before. It's not clear what exactly counts as a trial, in particular whether a trial may consist of moving coffee from one side to the other until the scale is balanced, but assuming that's valid, here's a way to do it in trials, using all three weights in ...

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I decided to try a different approach: Brute force. So, I created a Java 8 program to brute-force the creation of a decision tree in order to find how to weight the 12 coins in 7 measurements, and it worked. Further, I could use it to prove that this is the optimum, because it said that with 6 there is no solution. The program found a solution in 35 ...

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You can guarantee finding and determining the weight of the fake coin for Any useful weighing will be of an even number of coins, with the same number on each side of the balance. If the result is balanced, all the coins must be genuine and so will immediately disappear. Therefore if you start with an odd number of coins, there will always be the ...

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