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At the end of Probability class, our professor gave us the following puzzle:

There are 100 bags each with 100 coins, but only one of these bags has gold coins in it. The gold coin has weight of 1.01 grams and the other coins has weight of 1 gram. We are given a digital scale, but we can only use it once. How can we identify the bag of gold coins?

After about 5 minutes waiting, our professor gave us the solution (the class had ended and he didn't want to wait any longer):

Give the bags numbers from 0 through 99, then take 0 coins from the bag number 0, 1 coin from the bag number 1, 2 coins from the bag number 2, and so on until we take 99 coins from the bag number 99. Gather all the coins we have taken together and put them on the scale. Denote the weight of these coins as $W$ and the number of bag with gold coins in it as $N$, then we can identify the bag of gold coins using formula $$N=100(W-4950)$$ For instance, if the weight of all coins gathered is $4950.25$ grams, then using the formula above the bag number 25 has the gold coins in it.

My questions are:

  1. How does the formula work? Where does it come from?
  2. Do we have other ways to solve this puzzle? If yes, how?
  3. If the digital scale is replaced by a traditional scale, the scale like symbol of libra or the scale in Shakespeare's drama: The Merchant of Venice (I don't know what is the name in English), then how do we solve this puzzle?
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    $\begingroup$ If you can pull coins out of the bag, maybe you can see the difference instead of weighing. ;) $\endgroup$ – jpmc26 Nov 13 '14 at 19:54
  • $\begingroup$ @jpmc26 One of friends also said like that but our prof only smiled. :-D $\endgroup$ – Venus Nov 14 '14 at 9:54
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    $\begingroup$ This puzzle appeared on one of the episodes of Columbo: imdb.com/title/tt0075864 $\endgroup$ – Stephen Montgomery-Smith Nov 18 '14 at 23:04
  • $\begingroup$ @StephenMontgomery-Smith That's very old movie, I haven't been born yet when the movie aired so I don't watch it. Anyway, you're the one who also reviews Prof. Otelbaev's work, how is it going? Is it correct so far? $\endgroup$ – Venus Nov 19 '14 at 6:29
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    $\begingroup$ @Venus You can see that episode on Netflix. Otelbaev said there was a mistake in his work. math.stackexchange.com/questions/634890/… $\endgroup$ – Stephen Montgomery-Smith Nov 19 '14 at 14:46
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To understand the formula, it would be easiest to explain how it works conceptually before we derive it.

Let's simplify the problem and say there are only 3 bags each with 2 coins in them. 2 of those bags have the 1 gram coins and one has the 1.01 gram gold coins. Let's denote the bags arbitrarily as $Bag_0$, $Bag_1$, and $Bag_2$. Similarly to your problem, let's take 0 coins from $Bag_0$, 1 coin from $Bag_1$, and 2 coins from $Bag_2$. We know that the gold coins must be in one of those bags, so there are three possibilities when we weigh the three coins we removed:

Gold Coins in $Bag_0$: So the weight of the 3 coins on the scale are all 1 gram. So the scale will read 3 grams.

Gold Coins in $Bag_1$: So the weight of 1 of the coins is 1.01 grams and 2 of the coins are 2 grams. So the scale will read 3.01 grams.

Gold Coins in $Bag_2$: So the weight of 2 of the coins is 2.02 grams and 1 of the coins is 1 gram. So the scale will read 3.02 grams.

So each possibility has a unique scenario. So if we determine the weight, we can determine from which bag those coins came from based on that weight.

We can generalize our results from this simplified example to your 100 bag example.

Now for deriving the formula. Say hypothetically, of our 100 bags, all 100 coins in each of the 100 bags weigh 1 gram each. In that case, when we remove 0 coins from $Bag_0$, 1 from $Bag_1$, up until 99 coins from $Bag_{99}$, we'll have a total of 4950 coins on the scale, which will equivalently be 4950 grams. Simply put, if $n$ is our Bag number (denoted $Bag_n$), we've placed $n$ coins from each $Bag_n$ onto the scale for $n = 0, 1, 2, ... 99 $.

So the weight of the coins will be $Weight = 1 + 2 + 3 + ... + 99 = 4950$

But we actually have one bag with gold coins weighing 1.01 grams. And we know that those 1.01 gram coins must be from some $Bag_n$. In our hypothetical example, all of our coins were 1 gram coins, so we must replace the $n$ coins weighed from $Bag_n$ with $n$ gold coins weighing 1.01 grams. Mathematically, we would have: $Weight = 4950 - n + 1.01n = 4950 + .01n = 4950 + n/100$

Rearranging the formula to solve for n, we have: $100(Weight-4950) = n$, where $Weight$ is $W$ and $n$ is $N$ in your example.

I have no knowledge of an alternative answer to this puzzle, but perhaps another member's answer may be enlightening if there is. Technically speaking, you could have denoted the bags from 1 to 100 and gone through a similar process as above, but the method is still the same, so I wouldn't treat it as a new answer.

If our electric scale is replaced by a scale of libra, I don't believe it would be possible to answer this puzzle with only one measurement of weight. But again, perhaps another answer may be enlightening on that.

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  • $\begingroup$ To see why it's not possible to find the bag with just one "libra" weighing, consider that one weighing gives as 3 possible states (left, right, balanced), yet we have 100 bags to choose from. In other words we have a trit of information (trit: ternary equivalent of the bit) with each weighing. We will need at least 5 trits to distinguish between 100 possible bags. $\endgroup$ – Thanassis Oct 2 '18 at 5:29
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For #1: imagine for a moment that all the coins are fake. If we took 0 coins from bag 0, 1 coin from bag 1, 2 coins from bag 2... we'd have $99\times100/2=4,950$ coins, and those 4,950 coins would weigh a total of 4,950 grams. But now, say that bag 25 were the one with real coins that are slightly heavier: 0.01 grams heavier, in fact. So the total weight of the coins is $W=4950+0.01N$, where $N$ is the number of the bag with the real coins. But -- we have the weight, not the bag number. So let's invert the equation: we want to find N given W, not the other way around.

$$\begin{align}W&=4950+0.01N\\W-4950&=0.01N\\100(W-4950)&=N\end{align}$$

For #2, aside from renumbering the bags, there isn't a different way to do this; no matter what, we have to have a different number coming from the scale for each different possible result.

For #3, you need $\lceil\log_3k\rceil$ weighings to discover the odd coin out; for 100 coins, that's 5 weighings: the first splits the coins into groups of (up to) 34; the second into groups of (up to) 12; the third into groups of (up to) 4; the fourth into groups of (up to) 2; the fifth finds it guaranteed.

Why $\lceil\log_3k\rceil$? Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$.

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  • $\begingroup$ +1 for #3, but how to get this formula: $\lceil\log_3k\rceil$? $\endgroup$ – Venus Nov 13 '14 at 14:32
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    $\begingroup$ Each use of the balance scales actually compares three different groups of coins: the one on the left scale, the one on the right scale, and the one not on the scale at all. If one of the two groups on the scale is heavier, then the gold coin is in that group; if neither, then the gold coin is in the group not on the scale. Thus, each weighing can distinguish between 3 states, and $n$ weighings can distinguish between $3^n$ states. We need an integer solution to $3^n\ge k$, thus $n\ge\log_3k$, thus $n=\lceil\log_3k\rceil$ $\endgroup$ – Dan Uznanski Nov 13 '14 at 14:37
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According to the given values, the gold coins have essentially the same weight (down to a single percent) as the base ones. Since gold is heavy this means that each gold coin is significantly smaller.

Forget about weighing and just take the bag whose coins are much smaller than the coins in the other bags.

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    $\begingroup$ As much as this isn't the intended strategy, I really can't say this is wrong, because this would totally work. However, in the context of Math.se, it seems that this approach is a bit less mathematical than it was supposed to be. Perhaps it would be better suited for puzzling.se $\endgroup$ – Asimov Nov 13 '14 at 14:38
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    $\begingroup$ The fake ones could be made of tungsten; tungsten has a density remarkably similar to gold's. $\endgroup$ – Dan Uznanski Nov 13 '14 at 14:44
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    $\begingroup$ Perhaps the others are all gold-plated platinum or iridium, so they are very nearly the same size as the gold coins. $\endgroup$ – David K Nov 13 '14 at 14:44
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  1. If all coins would have equal weight ($1.00$ gram), then the total weight of taking $0$ coins from bag $0$, $1$ coin from bag $1$, etc. would be $4950$ grams. Verify: $1 + 2 + \cdots + 99 = 4950$. To work with the example: if you take $25$ coins from bag $25$, then the total offset in weight is $0.25$ grams.
  2. Not that I know of.
  3. When the rest of the rules are the same (weighing only once), you cannot solve the puzzle.
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    $\begingroup$ Re 3: With a balance that can only give three different "answers" (left is heavier, right is heavier, or equality), it takes at least five weighings (because $3^4<100$). $\endgroup$ – Hagen von Eitzen Nov 13 '14 at 14:26
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The total weight of coins on the digital scale is $$W=\sum_{m=0}^{99} mw_m$$ where $w_m$ is the weight of each coin in bag number $m.$ But if the gold coins are actually in bag number $N,$ then $$w_m = \begin{cases} 1 + 0.01 && \text{if}\ m = N, \\ 1 && \text{if}\ m \neq N. \end{cases}$$ Therefore the only term of the sum that is not equal to $m$ is the term $Nw_N,$ which is equal to $N + 0.01N.$ So $$W=\sum_{m=0}^{99} mw_m = \left(\sum_{m=0}^{99} m\right) + 0.01N = 4950 + 0.01N.$$ Knowing $W$, we solve for $N.$

Are there other solutions? Certainly! We could choose $m+1$ coins from each bag numbered $m$, or in fact any number of coins as long as we take a different number of coins from each bag and know which bag contributed what number of coins. The same calculation as before tells us how many gold coins are on the scale, and we then deduce which bag they must have come from.

In fact you can find the bag of gold coins among up to $101$ bags this way. That's because there are just $101$ different numbers of coins we can draw from a bag. If any of the bags have more coins, we can solve this for more bags. But we cannot solve the puzzle for so many bags that there would have to be two bags that contribute the same number of coins to the weighing, because if we found that that was the number of gold coins on the scale then we would not know which of the two bags they came from.

Now if you have only a two-pan balance that does not give a reading, it is no longer possible to determine where the gold coins are in one weighing. It will take multiple weighings, as in the solutions to this problem and this problem.

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