Finding the Fake Coin in Minimum Turns

[vikasj007]

One day i was given 9 gold coins, and told that one of them is fake and i had to find out which one is it, all i had was my two pan scale.

i was also told that all the coins looked identical, they were of the same shape and size, the only difference was that the fake was one gram lighter than the real ones(i was not told what is the actual weight of the real ones either). Now, using my two pan scale i had to find out the fake coin.


the real problem came when i was asked to do this in the minimum no. of turns.

Here i give you the same problem, and you have to tell me, how many turns do you need to find the fake, keep in mind that you have to do this in minimum no. of turns.

GOOD LUCK!

 

[prateek]

very easy
the answer is 2

make 3 piles of 3 coins. put 2 piles on 2 scales. keep the third one aside. if one of the two bundles on the scale is lighter, take that on, otherwise the third one. then put two of the correct pile on the scale, one on each pan. then the lighter one on the pan is fake, or if both of them are have same weight, the other one is fake.
correct?

 

[DarkEternal]

here's a much harder version. 12 coins, one of which is either heavier OR lighter than the others. all you have is a balance and 3 weighing trials. how do you determine which coin is different from the others?

answer: http://www.cut-the-knot.org/blue/weight1.shtml
but don't spoil it for yourself!

 

[FZ+]

Isn't this the same as the above?

First, two piles of 6. This let's you eliminate 6, and find out if it is heavier, or lighter.

Then do the same as prateek, with the pile of 6.

Or do you mean a top-pan balance?

 

[ceptimus]

No. You have to identify the one fake coin, and say whether it is heavier or lighter than the others, in just 4 weighings.

With two piles of six, one side of the weigh pan will go down, say the left one. Now you know either the fake coin is heaver, and on the left, or lighter and on the right. How does that narrow the search to six coins?

 

[DarkEternal]

yeah, the problem is quite a bit more difficult, as ceptimus noticed. the key point is you don't know if the fake coin is heavier OR lighter. by the way, you only have 3 trials, not 4.

 

[Parth Dave]

Divide the coins into 4 groups of 3. Place any two of the four onto the pan:
a. If they are equal, than you know the fake coin must be in one of the other two piles. Leave one of the two piles on the pan, replace the other pile with one of the remaining two piles. If the piles are again equal, than the last pile must have the fake coin in it. If they are not equal, than the one that was just placed onto the pan must have the fake in it.
b. If the original two are not equal, than you know that one of them has the fake coin in it. Also, the other two piles must have only real coins in them. Take off any pile, and replace it with one of the other two. If this time it is equal, than the pile that was taken off must have the fake coin in it. If they are not equal, than the pile that was left on the pan must have the fake in it.

Thus, in two turns we have found out the coin is one of three coins.
Put any two of the three coins onto the pan. If they are equal, than the last coin must be fake. If they are not equal, than one of the two on the pan are fake. Take off any of the two, and replace it with the last one. If these coins are equal, than the coin taken off is fake. If they are not equal, than the coin that stayed on is fake.

 

[DarkEternal]

Divide the coins into 4 groups of 3. Place any two of the four onto the pan:
a. If they are equal, than you know the fake coin must be in one of the other two piles. Leave one of the two piles on the pan, replace the other pile with one of the remaining two piles. If the piles are again equal, than the last pile must have the fake coin in it. If they are not equal, than the one that was just placed onto the pan must have the fake in it.
b. If the original two are not equal, than you know that one of them has the fake coin in it. Also, the other two piles must have only real coins in them. Take off any pile, and replace it with one of the other two. If this time it is equal, than the pile that was taken off must have the fake coin in it. If they are not equal, than the pile that was left on the pan must have the fake in it.

Thus, in two turns we have found out the coin is one of three coins.
Put any two of the three coins onto the pan. If they are equal, than the last coin must be fake. If they are not equal, than one of the two on the pan are fake. Take off any of the two, and replace it with the last one. If these coins are equal, than the coin taken off is fake. If they are not equal, than the coin that stayed on is fake.

not a bad solution, except you use too many trials in the last case. only 3 trials allowed total!

 

[ceptimus]

Warning! Spoiler follows...












Label the coins 1 to 9, A, B, C

perform 3 weighings X, Y and Z:

X: 1234 v 5678
Y: 1489 v 36AB
Z: 278A v 46BC

for weighings X, Y, Z there are three possible outcomes:
= Balance
< Left pan goes down
> Right pan goes down

If the fake coin is F, we expect the following result:

F Result

1 <<= or >>=
2 <=< or >=>
3 <>= or ><=
4 <<> or >><
5 >== or <==
6 >>> or <<<
7 >=< or <=>
8 ><< or <>>
9 =<= or =>=
A =>< or =<>
B =>> or =<<
C ==> or ==<

The result not only identifies the fake, but tells you whether it is heavier or lighter than a good coin.

 

[vikasj007]

ceptimus, i think you are on the right track, and you should try this in the next puzzle (fake coin#2), maybe you will have any luck there as i don't think anybody else is goin to...