# You are given twelve gold coins and a simple balance scale. One of the coins is counterfeit.?

You are given twelve gold coins and a simple balance scale. One of the coins is counterfeit, but it is not known whether it is lighter or heavier than the other coins. You must find it, using only three weighings with the scale.

This is quite hard. You must consider & show all possibilities to get 10 more points . Give it a try…!

Remember, ONLY in THREE turns….!!

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I believe you want the counterfeit coin identified in 3 weighings, not just whether it is lighter or heavier as the other answerers assumed.

An interesting problem. Takes a bit of thought, but it can be done in 3 weighings.

In the following "X" stands for the counterfeit coin

First divide the 12 coins into 3 groups of 4.

First weighing: Place one group of 4 on each side of the balance.

Case 1: they balance

If the two groups balance then the unweighed group of 4 contains X. For the 2nd weighing, take two of the 4 coins that have not been weighed and put them on the scale. If they balance then remove one of these coins and place one of the remaining two coins on the scale. If these two balance then the unweighed coin = X (note: this is the only time you will have identified X but know if it’s lighter or heavier). If they don’t balance then the coin last placed on the scale =X (and we’ll know if it’s heavier or lighter from the direction of the tilt). If the original two placed on the scale don’t balance then one of them is X. Remove one and replace it with a normal coin and if they balance, then the removed coin = X, and if they don’t balance then the coin that wasn’t removed = X.(And again we’ll know if X is heavier or lighter from the direction of the tilt).In any case 2 more weighings (3 total) suffices.

Case 2: the two groups of 4 don’t balance.

In this case label the coins that are on side one(S1):1a, 1b, 1c and 1d; and other side, S2: 2a, 2b,2c and 2d.

So the two sides of the scale are as follows:

1a 2a

1b 2b

1c 2c

1d 2d

S1 S2

The scale will be tilted in one direction.

We know the ones that haven’t been weighed are normal.

We will use these normal ones as replacements "R".

For the second weighing we do the following: We remove 2a, 2b and 2c and replace them with 1a,1b and 1c. And we substitute normal coins (R) in place of 1a,1b and 1c

So, for the second weighing the scale looks as follows:

R 1a

R 1b

R 1c

1d 2d

S1 S2

Possibility 1: the scale is now balanced.

Now, if the scale balances then one of 2a, 2b,2c =X, but also note that we would now know if X is lighter or heavier than the others, since if side 2 had been tilted down before balancing then X would be heavier, and if side 2 had been tilted up before balancing then X would be lighter.This allows us to determine which of 2a,2b,2c = X with one more weighing. Put 2a and 2b on the scale and if they balance 2c =X. If it tilts, then we know which one it is since we know if X is heavier or lighter. So one more weighing (3 tolal) suffices.

Possibility 2: The scale remains tilted in the same direction.

If the scale stays tilted in the same direction then one of

1d, 2d = X since if one of 1a,1b, 1c = X then the scale would tilt in the opposite direction since they have been switched to the other side. Thus, if we put 1d on the scale with a normal coin, and it tilts, then 1d=X and if it doesn’t then 2d=X. So, again 3 weighings suffice.

Possibility 3: The scale tilts in the opposite direction

If the scale tilts in the opposite direction then one of a1, b1 or c1 = X, and as in possibility 1 (Case 2), we now know whether X is heavier or lighter, and the analysis is the same as it was for possibility 1; and 3 weighings suffice.

So it can be done in 3 weighings.

label the coins with the letters from FAKE MIND CLOT

and weigh the coins: MA DO — LIKE, ME TO — FIND, FAKE — COIN. Logic will now suffice to find the odd coin. For instance, if the results are left down, balance, and left down, then coin "A" is heavy.

well…

first, if the coin is counterfeit, it must be lighter not heavier. any metal heavier than gold would be too valuable to waste as a counterfeit coin for gold..so a lighter coin would indicate the counterfeit.

place six coins each on the scale. the scale side that is lighter is the one with the fake coin.

divide that lighter side to 3 coins for each arm. again, the lighter side contains the fake.

from these coins, weigh just two:

if both coins are equal, the unweighed coin is the fake.

if you again have a lighter coin, the lighter coin is the fake.

all done with three weighing…based on the premise that the material used in the fake cannot be lighter because items heavier than gold are more or as valuable than gold…so an inferior weighted metal must be substituted.

Bear with me.

Label the coins A-L

If we place A,B,C,D,E & F on one side and G,H,I,J,K & L on the other side we have equal piles of 6 coins

Now place A, B, C & D on left side of the scale and G, H, I & J on the right side. View results. If both sides are equal, then the fake coin is not on the scale. If the sides are not equal then the four not on the scale, (E, F, K or L), are normal.

Assuming the eight coins are normal, place E & F on the left side and K & J on the right. (Remember J is a normal coin). Now, if they balance out, we know that coin L is abnormal. Balance coin L against coin J to find if the counterfeit one is light or heavy.

Now, if the second weighing determines that K & J is heavier than E & F, we know that either K is too heavy or E is light or F is light. Place E on one side and F on one side. If they are equal, then K is heavy. Otherwise, we will find that E or F is light based on the scale.

Going back. If in the first weighing the side G, H, I & J is heavier than side A, B, C & D, either A,B,C & D is light or G,H, I & J is heavy. Take A, B & G on one side and C, H & E on the other. (Remember that E is normal because the first time weighed, it was not on the scale). If they are equal, then either D is light, or I is heavy or J is heavy. By placing I on one side and J on one side, we get our answer. If they balance, D has to be light, but if they do not balance, the heavier one is the fake one.

When we weigh A, B & G against C, H & E and find the right side is heavy, then either H is heavy, or A is light or B is light. By placing A on one side and B on the other we determine as above.

But if we weigh A, B & G against C, H & E and the right side is lighter, then either C is heavy or G is light. By weighing C against E (the good coin) we get our answer of too light or too heavy.