# Tag Info

## Hot answers tagged weighing

49

There is another way to do this problem, that doesn't involve any sort of conditional branching at all. It is in fact possible to set a fixed weighing schedule beforehand and still determine which ball is lighter or heavier in just 3 weighings. I'll explain how below. The gist of problems like these is, how much information can you get from the procedure ...

43

The solution is simple:

42

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 ...

31

Split this into three groups of four, A1, A2, A3, A4; B1, B2...; C1, C2... Each step here corresponds to one weighing. Weigh A against B. If A > B, then weigh A1, B1, and B2 against B3, B4, and C1. If the weights are equal, then one of A2...4 is heavier; weigh A2 and A3. If they are equal, A4 is heavier. If one is heavier, then that ball is heaviest. If ...

26

You can number stacks from 1 to 10. From each stack take as many coins as the stack number. Weigh them all. Subtract the weight from 10+20+...+100=550. The result is the number of the silver stack. P.S. You can do the same even if you have 11 stacks of 10: just number them from 0 to 10.

25

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 ...

24

It turns out that you need four weights, measuring 1, 3, 9, and 27 pounds. The trick here is that you can put weights on both sides of the scale. If you have an object of 2 pounds on the left side, you can place the 3-lb. weight on the right side and the 1-lb. weight on the left side for the scale to balance. Similarly, if the object weighs 5 pounds, you ...

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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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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 ...

21

Yes. Solution:

19

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 ...

17

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 ...

15

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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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 ...

14

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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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 ...

14

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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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 ...

13

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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The answer is I ran a test with 6 coins, leaving 2 off each time. I got that to work. This made me think I could scale it up to 10 coins by doing 3 on each side. I tried that and got all the cases where 2 coins were different, but 3 coins would stop me (and I thought only the even sets would be a problem). That made me try a different number of coins in ...

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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 ...

12

An obvious way to select the weights is to have them be powers of 2, from 1 to $2^{99}$. Every positive integer has a unique binary representation, so no two different sets of weights can be have the same total weight. There are $2^{100}$ possible sets of weights. If any two of them have the same weight, then they can be balanced (removing any weights that ...

12

Let's try looking at it from the biggest weights first. Let $w$ be the biggest weight of one such system, and let $n$ be the number of $w$-weights in the system. Let the remainder $r = 212 - n \times w$. Then $n$ is maximal (subject to $n \times w \leq 212$) and $r = w-1$. Indeed, if $n$ were not maximal, then $r \geq w$. We can do division with ...

12

There are $5!=120$ possible orderings of the fencers, so we need $\log_2(120)\approx 6.9069$ bits of information. Each duel provides at most $1$ bit of information, so at least $7$ duels will be necessary. Here is a way to determine the ordering in $7$ duels. Call the fencers $A$, $B$, $C$, $D$, and $E$. First $A$ and $B$ duel, by symmetry we assume $A$ ...

12

You are looking for 10 numbers such that no two subsets of the numbers sum to the same amount, such that the largest number is minimal. This is conjectured to correspond to sequence A005318 in OEIS, which gives the 10th element as 309. A096858 provides the corresponding amount of coins from each pile as 148, 225, 265, 285, 296, 302, 305, 307, 308, and 309....

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Firstly, use the balance scale once to compare the weights of If $A_1$ and $B_1$ both weigh the same, then both swapped eggs are in the same one of these two sets. In that case, use the balance scale again to compare the weights of If $A_2$ and $B_2$ both weigh the same, then the two swapped eggs must both be in the same one of $A_1,B_1$ and also both in ...

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And now my computer generated and checked solution: Put the weights in a row and number the places from 1 to 15. Then do the following: I generated these command by an similar algorithm as Murch. But with the difference if I compare three weights at $\left(\begin{array}{c}x_1\\y_1\end{array}\right)$, $\left(\begin{array}{c}x_2\\y_2\end{array}\right)$, \$\...

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