To Find Fake Coin

In February 2003, Harold Hopwood of Gravesend wrote a short letter to the Daily Telegraph, saying that he had solved the newspaper’s crossword every day since 1937, but one conundrum had been nagging away at the back of his mind since his schooldays, and at the age of 82 he had finally decided to enlist some help.

The puzzle was this. You are given 12 coins. They all have the same weight, except for  one, which may be either lighter or heavier than the rest. You have to find out which coin is different, and whether it is light or heavy, using at most three weightings on a pair of scales. The scales have no graduations for weights; they just have two pans, and you can tell whether they are in balance, or the heavier one has gone down and the lighter one has gone up.

Before reading on, you should have a go. It’s quite addictive.

Within days, the paper’s letters desk had received 362 letters and calls about the puzzle, nearly all asking for the answer, and they phoned Ian Stewart. He recognized the problem as one of the classic puzzles, typical of the weights and scales genre, but he had forgotten the answer. But his friend Marty, who happened to be in the room when he answered the phone, also recognized the problem. The same puzzle had inspired Marty as a teenager, and his successful solution had led him to become a mathematician.

Of course, he had forgotten how the solution went, but they came up  with a method in which they weighed various sets of coins against various others, and faxed it to the newspaper.

In fact, there are many answers, including a very clever one which Ian Stewart finally remembered on the day that the Telegraph printed out less elegant method. Professor Stewart had seen it twenty years earlier in New Scientist magazine, and it had been reproduced in Thomas H. O’Beirne’s Puzzles and Paradoxes, which Professor Stewart had on his bookshelf.

Puzzles like this seem to come round every twenty years or so, presumably when a new generation is exposed to them, a bit like an epidemic that gets a new lease of life when the population loses all immunity. O’Beirne traced it back to Howard Grossman in 1945, but it is almost certainly much older, going back to the seventeenth century. It wouldn’t surprise me if one day we find it on a Babylonian cuneiform tablet.

O’Beirne offered a ‘decision tree’ solution, along the lines that Marty and Professor Stewart had concocted. He also recalled the elegant 1950 solution published by “Blanche Descartes” in Eureka, the journal of Archimedeans, Cambridge University’s undergraduate mathematics society. Ms Descartes was in actuality Cedric A. B. Smith, and his solution is a masterpiece of ingenuity. It is presented as a poem about a certain Professor Felix Fiddlesticks, and the main idea goes like this:

F set the coins out in a row

And chalked on each a letter so,

To form the words “F AM NOT LICKED”

(An idea in his brain had clicked.)

And now his mother he’ll enjoin:


This cryptic list of three weightings, one set of four against another solves the problem, as Eureka explains, also in verse. To convince you, Professor Stewart listed all the outcomes of the weightings, according to which coin is heavy or light. Here, L means that the left pan goes down, R that the right pan goes down, and — means they stay balanced.

\begin{tabular}{|| l | l | l | l ||} \hline    False coin & 1st weighing & 2nd weighing & 3rd weighing \\    F heavy & --- & R & L \\    F light & --- & L & R \\    A heavy & L & --- & L \\    A light & R & --- & R \\    M heavy & L & --- & L \\    M light & R & R & --- \\    N heavy & --- & R & R \\    N light & --- & L & L \\    O heavy & L & L & R \\    O light & R & R & L \\    T heavy & --- & L & --- \\    T light & --- & R & --- \\    L heavy & R & R & R \\    L light & L & L & L \\    C heavy & --- & --- & R \\    C light & --- & --- & L \\    K heavy & R & --- & L \\    K light & L & --- & R \\    E heavy & R & L & L \\    E light & L & R & R \\    D heavy & L & R & --- \\    D light & R & L & --- \\ \hline    \end{tabular}

You can check that no two possibilities give the same results.

The Telegraph’s publication of a valid solution did not end the matter. Readers wrote in to object to Prof Stewart and his colleagues answer, on spurious grounds. They wrote to improve it, not always by valid methods. They e-mailed to point out Ms Descartes’s solutions or similar ones. They told them about other weighing puzzles. The readers thanked the duo for setting their minds at rest. The readers cursed the duo for reopening old wounds. It was as if some vast secret reservoir of folk wisdom had suddenly been breached. One correspondent remembered that the puzzle had featured on BBC television in the 1960s, with the solution being given the following night. Ominously, the letter continued, “I do not recall why it was raised in the first place, or whether that was my first acquaintance with it; I have a feeling that it was not.”

More puzzles for you later,

Nalin Pithwa


  1. Anubhav C. Singh
    Posted August 25, 2015 at 4:11 pm | Permalink | Reply

    So the drawback with this logic is that it will only work when you know if the specific coin you are searching for is heavier or lighter than the rest. Anyway, here it is :-
    Let us assume that the coin is heavier than the rest.
    Split the coins into 3 groups of 4.
    Now weigh two of the groups on the pans. If one turns out to be heavier, chose that one. Otherwise if the pans are is equilibrium, chose the third.
    Now you have 4 coins and two weighings more. So split them into two each, find the heavier group, and weigh the last two members of that group to find the special one.
    However, this logic falters when you don’t know how the coins differ in the first step itself; as one pan will go up, another will go down. To verify this we need to weigh one of the groups with the last group. If the weights are still unequal, the group you chose is the odd one, while if they are in equilibrium, the one you excluded is the one containing the odd coin.
    If the groups are equal in the first step itself, the excluded one is the odd one, but you will still have to weigh the excluded one to deduce whether the special coin is heavier or lighter.
    In any case, the count of usage of chances will go to 4, so feel free to make any modifications to my logic as you like.

    • Posted August 26, 2015 at 1:48 pm | Permalink | Reply

      excellent attempt! Keep up the good work of thinking! Do you know that IBM logo is THINK? Well, I will put my answer separately now…

  2. Pradeep Selva
    Posted October 9, 2015 at 10:58 am | Permalink | Reply

    So here we are given 12 coins….. so we split them into two ……so we will have 6 coins in each group. if we put them on a balance then 1 will be heavier or lighter for sure. Then split each set . So now you will be left with 4 groups of 3. Now take 2 groups separately.Now put 2 groups on the balance. if they are equal take them away and if they are unequal keep them. Now take the remaining 2 groups and place them. They must be unequal. so take each set and with each set do the following-
    1. place 2 random coins on the balance
    * if they are equal, then place one of the coins on the balance and place the remaining coin in the group on the other pan. if they are equal then that is not the set we are looking for.
    * if they are unequal then that is the set we are looking for….So leaving any one coin on the pan and take the other away and place the remaining coin of the group on the other pan.
    => If they are equal , then the coin that we just took out of the pan is the fake one.
    => If they are unequal, then the coin that we just left in the pan is the fake one.

    • Posted October 10, 2015 at 1:35 pm | Permalink | Reply

      will reply later; good attempt Pradeep

      • pradeepselvaraju
        Posted October 13, 2015 at 10:39 am | Permalink

        Okay sir…

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