You are, as you have been many times before, in some place that has a balance scale and 12 coins.

11 are all "normal" and stay normal, but one of these coins, however, is shifty -- as soon as you "finish" a weighing by taking the coins off the scale, it changes its weight! (This transition cannot be seen but you know that it happens)

To elaborate, any time you use any set of coins in a weighing the shifty coin shifts its weight. This applies even if the shifty coin is set aside during the weighing.

It switches its weight in the order Lighter-Heavier-Normal, but to add to your difficulty, you do not know what state it is in at first.

In how many weighings can you spot the shifty coin?

  • $\begingroup$ A virtual scale will come in handy. $\endgroup$ Nov 26, 2020 at 16:59
  • 6
    $\begingroup$ the weight of the shifty coin changes every time you put it in the scale or every time you use the scale even without shifty one on the scale? $\endgroup$
    – Oray
    Nov 26, 2020 at 20:08
  • $\begingroup$ Edited to elaborate. $\endgroup$
    – Andrew
    Nov 27, 2020 at 10:31

2 Answers 2


Assumption: Shifty coin only switches if weighed.

I believe this is optimal solution;

Weigh 6v6

if (case 1)

it is in balance, that means the shifty one was in the state as normal and next time it will act as lighter.

otherwise (case 2)

no balance, it could be heavier or lighter. Next time it could be heavier or normal.

If it is Case 1;

weigh again 4v4 (the rest is not weighed);

if (Case 1.1 - They are equal)

that means our shifty weight is in the not weighed 4 coins and we know that it will be lighter next time we weigh. You can easily tell that it will require two more weighing at most since it will be lighter than heavier 2v2 etc. (4 weighs required)

if (Case 1.2 - They are not equal)

that means lighter 4 coin side have t consist our shifty one, from now on, shift coin will act as heavier.


weigh again 1v1 in that group of coins, if one of them is heavier, it is our coin, if they are equal weigh the rest 1v1 and whichever is heavier is our coin (4 weighs required)

if Case 2 happens (no balance at first weighing);

our coin could be anywhere, but if we weight 6v6 again to be sure whether if it was lighter or heavier, this would be a problem for our next weighs and only one information will be gained, so we switch 3 coins between sides and weigh again;

if there is balance (Case 2.1), (also meaning it was heavier at first)

the shifty coin on the heavier side of our original weighing (we know that coin will act lighter now), then take these original 6 (not switched ones) and weight 2v2, if there is balance, weigh 2 coins which were not weighed and find our shify coin, if one side was lighter, our coin is there weigh again and find our coin (4 weighs required). If there is no balance, our coin in the lighter part and will act as heavier next time, weigh 1v1 and choose heavier one.

if no balance again as the beginning and the same side is lighter again (Case 2.2) (that also means our coin was lighter at first weighing.)

our coin is in the 3-switch coins and

as in table below;

enter image description here


it was not in the switch ones

as in the table below;

enter image description here

so we know that

our coin is in one of the 3 coin-pack and we know which 3-coin pack consists of our shifty one, but one weighing will be waste for sure as we know that it will act normal since it was heavier previously, take one normal coin and weigh 2v2, then take that normal coin out (our shifty coin will act as lighter now) and weigh 1v1, if there is balance, our fake is not weighed one, if one side is lighter, that coin is the shifty one.

so the answer is


  • $\begingroup$ I think you are missing that the weight changes after each weighing, e.g. in case 1 the 4th weighing does not give information since the shifty coin is back to 'normal'. $\endgroup$
    – Retudin
    Nov 26, 2020 at 20:03
  • $\begingroup$ OK I assume we'll hear if your interpretation/answer is ok. $\endgroup$
    – Retudin
    Nov 26, 2020 at 20:20
  • 1
    $\begingroup$ This may not be the solution to the puzzle as OP posed it, but it is the solution to the best possible version of this puzzle. $\endgroup$
    – Bass
    Nov 26, 2020 at 23:56
  • $\begingroup$ Congratulations! Although this was not what I intended, it is still a clever solution and you probably put a lot of effort, so here's an upvote. $\endgroup$
    – Andrew
    Nov 27, 2020 at 10:32

With the conditions stated in the problem, you cannot possibly find the shifty coin in less than

5 weighings

This is because

You can't get any information from the weightings when the shifty coin has normal weight. The best possible thing you can do is find out the state of the coin during that weighing if you didn't know it already.

This means that

If the coin starts in normal state, the first weighing is useless. Then you have 2 weighings to find the coin - impossible out of 12 coins. The fourth weighing the coin is normal again and is useless. Finally on the fifth weighing you can find the coin.

Since you can't do better than what I've stated, it's easy to find a solution, there are many permutations, here's an example:

Weigh 6 vs 6. If one side goes down, weigh the lighter side 3 vs 3. If one side goes down that means the coin started light and is now heavy, it will be normal next so weight whatever, on the fourth weighing it will be light, weigh the heavy side 1 vs 1 and find the coin.
If the 3 vs 3 weighing was balanced, that means the coin was heavy first and is now normal, it will be light next. Take the heavy 6 coins from the first weighing and weight 3 vs 3, next weight the light side 1 vs 1, the heavy coin is shifty.
If the first weighing was balanced that means the coin started normal. Weigh 6 vs 6 again, then weight the light side 3 vs 3, the fake coin is on the heavy side, however the fourth weighing it will be normal again, weigh whatever, finally on the fifth weighing weigh the heavy side 1 vs 1 and find the coin - it will be light.


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