Can You Solve The Counterfeit Coin Riddle? – Jennifer Lu

16 thoughts to “Can You Solve The Counterfeit Coin Riddle? – Jennifer Lu”

  1. I don’t get it…if the king already knows that the fake weights different ..he could have picked it by multiple weighing … I mean common no big deal weighing 12 damn coins!

  2. I figured a slightly different way.
    I take six and six and weight them. The one that is lighter I divide into two piles of three that I than again weight against one another. The light pile I seperate into all three. Two of which I weight. If one is lighter I know its the counterfeit, if they are both equal than I know I am holding the false coin.

  3. take 2 piles 6 and 6 put aside take half that is heavier of the new lighter 6 take 3 and 3 weigh them put heavier pile aside take last 3 weigh 2 of them if they are even last one is thrown out if one is lighter throw it out same answer no marker xD

  4. There is an easier solution. Here it goes.

    1. Take 6 coins, and weigh against the remaining 6 to know which pile contains the fake one.
    2. Now from the 6 coins which are light in weight, take 3 coins and compare them against the remaining 3.
    3. Now we are down to 3 coins in which 1 is fake. So we take out 1 coin at random and weigh the remaining 2 coins against each other. If they are equal, the 1 coin taken out is the fake. If any of the coins on the weigh scale is light, you will be easily able to see it.

  5. we can do the same with 2-2 painting and swapping the coins on first unbalanced 4-4 weigh and…voila….!!

  6. just use your hands to weigh them, 2 at a time. he only gave you a limit to how many times u can use the scale. he never gave u a limit on how much u could use anything else

  7. Why does the king need the greatest math Magician if someone normal could just keep using the scale the regular way and find the fake coin?

  8. I doesn’t matter, if they look the same then you can pick one and you’ll have the same amount of coins in circulation.

  9. I did this but instead of 3 groups of 4 I did 2 groups of 6 then moved to groups of 3, single group of three, then if balanced remaining coin is fake.

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