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Probably a very easy puzzle for you guys; my point is more a mathematical one.
FYI, it was shown in an episode of Columbo.

You have 3 bags of coins. Each bag contains 50 identical coins. Two bags contain genuine gold coins, weighing 1 lb each (yes, I know, but that's what Columbo said!), one bag contains fake coins, weighing 1 lb 1 oz each, but for the rest completely identical in appearance to the genuine ones.
The task is to find which bag contains the fake coins, doing a single weighing on a scale with 1 plate (so no, you can't put one bag on one side and one bag on the other).

The solution given by Columbo was the following:

Number the bags from 1 to 3, take 1 coin from bag 1, 2 coins from bag 2 and 3 coins from bag 3. Put all 6 coins together on the scale. If the weight is 6 lb 1 oz, bag 1 contains the fake coins, if it is 6 lb 2 oz, bag 2, if it is 6 lb 3 oz, bag 3.

My doubt is:

Is this method general, meaning could you do it with any number of bags? And in that case would you have to adjust the number of coins taken from each bag? And does that depend on the difference in weight between fake and genuine coins?
The total weight would be something like: $W = \sum_{i}^{N_{bags}} {n_i \cdot w_i}$, where $n_i$ is the number of coins taken from bag $i$ and $w_i$ is the weight of 1 coin in bag $i$.
If only one bag (say the one numbered j) contains fake coins, the sum can be rewritten as: $W = n_j \cdot w_F + (\sum_{i \ne j}^{N_{bags}} {n_i}) \cdot w_G$, where $w_F$ is the weight of a fake coin and $w_G$ is the weight of a genuine coin.
And considering that: $dw = w_F - w_G$, $W$ can be further rewritten as: $W = n_j \cdot dw + (\sum_{i}^{N_{bags}} {n_i}) \cdot w_G$.
So I suppose Columbo's method only works when there is only one distinct value of $W$ for each possible $j$, which I assume can be imposed by an appropriate choice of the values of $n_i$ (although I would not know how).

What do you think? Is there any post or other resource where this theory is discussed?

Thanks!

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    $\begingroup$ You’ll need to state the conditions of your generalisation. Columbo’s method works with any number of bags with identical coins per bag, with exactly one bag containing fake coins, provided you have enough coins to put on the scale. It doesn’t work if a bag can contain different coins, or if you have multiple fakes where different linear combinations produce the same sums, etc. $\endgroup$
    – Lawrence
    Nov 3, 2019 at 14:02
  • $\begingroup$ You're right! I had not considered that when $n_i = i$ the second term in the sum is a constant depending only on $N_{bags}$, thus if $n_j = j$, $W$ is indeed different for each $j$. $\endgroup$
    – user6376297
    Nov 3, 2019 at 14:51
  • $\begingroup$ The only thing Columbo needs to look out for here is when there are less coins than bags - so, say if there were only two coins in each bag, Columbo's method doesn't work. $\endgroup$
    – JMP
    Mar 10, 2020 at 10:56
  • $\begingroup$ Given that you know there is exactly one bag with fake coins, you can omit one bag from the testing. Take one coin from bag one and two coins from bag two. If you get three pounds exactly, the third bag has the fake coins. $\endgroup$
    – Jason Goemaat
    Oct 23, 2020 at 18:20

3 Answers 3

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Colombo’s method works like this:

  1. Label the bag with number 1, 2, and 3.
  2. Take $i$ coin(s) from bag #$i$ and put them all on a scale.
  3. You will get $6 lb + k oz$. Here bag #$k$ is the one containing fake coins.

This method only works if there is only one bag containing fake coins.

My method will be:

  1. Label the bags with number 0, 1, 2.
  2. Take $2^{i}$ coin(s) from bag #$i$ and put them all on the scale.
  3. You will get $7 kg + \sum_{i=0}^{2} a_{i}2^{i} oz$ where $a_{i}$s are either 0 or 1.

If $a_{i}=1$ then bag #$i$ contain fake coins.

For example if you get $7kg + (2^{0} + 2^{2}) oz$, then bag #$0$ and #$2$ contain fake coins.

Note: This method works for any number of bags, provided there are enough coins in each bag. This method can detect how many and which bags contain fake coins.

If you are curious why is this the case, You may want to google why power of 2 spans integer numbers.

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  • $\begingroup$ The problem with the generalizations in practice is that there are error bars on all the weights, and a relatively small error bar could easily shift your results. $\endgroup$
    – Dr Xorile
    Jan 3, 2022 at 5:34
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“Is this method general, meaning could you do it with any number of bags?”

Yes, it’s possible with any number of bags depending on the amount of coins in each bag. If you want to investigated N bags, using the method described in Columbo (thus: only 1 bag with fake coins and taking 1 coin from the first bag, 2 coins from the second etc.), the following needs to hold:

Bag 1 contains >=1 coin

Bag 2 contains >=2 coins

Bag 3 contains >= 3 coins

….

Bag N-1 contains >= N-1 coins

Bag N contains >= N coins

“And in that case would you have to adjust the number of coins taken from each bag?”

No, as described above this is not necessary. Only the amount of coins in the bags need to comply to minimum amount of coins.

“And does that depend on the difference in weight between fake and genuine coins?"

No, as long if the difference in weight between the fake and genuine coin is known this doesn't matter. Assuming the conditions for minimal amount of coins in a bag described above are satisfied, only one bag contains fake coins and the solution of Columbo is used.

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Put the three bags on the scale. Take one coin from one bag and put it on your right side, put a mark on that bag, take one more coin from a different bag and put it on your left side. By so doing you know which bag contains the fake coins.

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