# You have 12 coins, one of which is fake.

Hey friends, try and solve this intricate puzzle. You can use your skills to explain this puzzle using a flow chart. If you make one send it to me on my email address, I will include it in the body of this post itself.

## 12 coins 3 weighings

You have 12 coins, one of which is fake. The fake coin is indistinguishable from the rest except that it is either heavier or lighter, but you don’t know which. Can you determine which is the fake coin and whether it is lighter or heavier using a balance scale and only 3 weighing?

18 Comments

ok if a coin being heavy or light is not known before…dan if we compare 5-5 and they being equal, compare remaing two..mark heavier coin as H and lighter one as L..in 3rd chance..compare either..H or L with…one coin from remaining 10….IF U TAKE “L” AND IT BALANCES..THAN FAULT COIN IS HEAVIER AND IS “H” ELSE FAULT COIN IS “L” AND IS LIGHTER

initially weight 5 to 5 and keep 2 aside…..if both sides are equal…the heavier ball is in the remaining 2 and in nxt chance u can find dat..here only 2 chances are used to find heavier coin :p

if heavier coin is in one of 5 set of coins..now take 2-2 pair n compare..if they are equal…the reamaining coin is heavier else…in the nxt chance u can compare..remaining 2 over n done !

it is so hard to think .wait let me try .got it!.there is no such thng as fake coin

govindarajan,yes that is correct

I for one is of the opinion that school children must be exposed to this logical thinking.

i think minimum 4 weightings r required… Rajesh’s method is d shortest but the problem is that the fake coin may b lighter..

no vivek,the thing is that the fake coin must b heavier

Number the coins 1 through 12.

1. Weigh coins 1,2,3,4 against coins 5,6,7,8.

1.1. If they balance, then weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin).

1.1.1. If the second weighing also balances, we know coin 12 (the only one not yet weighed) is the counterfeit. The third weighing indicates whether it is heavy or light.

1.1.2. If (at the second weighing) coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light. Weigh 9 against 10. If they balance, 11 is heavy. If they don’t balance, you know that either 9 or 10 is light, so the top coin is the fake.

1.1.3 If (at the second weighing) coins 11 and 8 are lighter than coins 9 and 10, either 11 is light or 9 is heavy or 10 is heavy. Weigh 9 against 10. If they balance, 11 is light. If they don’t balance, you know that either 9 or 10 is heavy, so the bottom coin is the fake.

1.2. Now if (at first weighing) the side with coins 5,6,7,8 are heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy. Weigh 1,2, and 5 against 3,6, and 9.

1.2.1. If (when we weigh 1,2, and 5 against 3,6 and 9) they balance, it means that either 7 or 8 is heavy or 4 is light. By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit.

1.2.2. If (when we weigh 1,2, and 5 against 3,6 and 9) the right side is heavier, then either 6 is heavy or 1 is light or 2 is light. By weighing 1 against 2 the solution is obtained.

1.2.3. If (when we weigh 1,2, and 5 against 3, 6 and 9) the right side is lighter, then either 3 is light or 5 is heavy. By weighing 3 against a good coin the solution is easily arrived at.

1.3 If (at the first weighing) coins 1,2,3,4 are heavier than coins 5,6,7,8 then repeat the previous steps 1.2 through 1.2.3 but switch the numbers of coins 1,2,3,4 with 5,6,7,8.

Let us give numbers to the coin as 1,2,3,4,5,6,7,8,9,10,11,12

Put coin numbers 1,2,3,4 on one side of scale and 5,6,7,8 on other side of scale. If both are equal then defective coin is in the group of coin numbers 9,10,11,12.

Now put coin number 1,2 on one side of scale and coin number 9,10 on other side of scale. If they are equal then defective coin is 11 or 12

Now put coin number 1 on one side of the scale and coin number 11 on other side. If they are same than coin number 12 is defective.

If on third weighing it defers than coin number 11 is defective.

Deepak, in ur soln. in 4th step, if one group weighted wit 3gud coins r of equl size it means dat d last left out 3 coins has d defective… in dis case u nvr knw wthr d defective coin s lighter r heavier.. Den hw 2 find d defective 1 from those 3coins.. ?????????

Preparation: Divide the 12 coins into 3 groups of 4 coins each say A,B,C.

Divide each group into subgroups of 1 coin and 3 coins.

Group A is split as A1, A3, group B as B1, B3, and group C as C1 and C3.

First weighing: Place A1, A3 on left pan, B1, B3 on right pan. Note the condition of balance.

Second weighing: Rotate A3 to right pan, B3 out of pan and C3 to left pan.

Third weighing:

Case a. If there is no change in balance condition – all 3 coin groups are good coins. Clear them from the scene and in third weighing find out which single coin is of odd weight.

Case b. If there is change in balance condition – you know which of the 3 coin groups is containing odd coin. Clear all other coins from the scene and in third weighing find which of the 3 coins is of odd weight.

Preparation: Divide the 12 coins into 3 groups of 4 coins each say A,B,C.

Divide each group into subgroups of 1 coin and 3 coins.

Group A is split as A1, A3, group B as B1, B3, and group C as C1 and C3.

First weighing: Place A1, A3 on left pan, B1, B3 on right pan. Note the condition of balance.

Second weighing: Rotate A3 to right pan, B3 out of pan and C3 to left pan.

Third weighing:

Case a. If there is no change in balance condition – all 3 coin groups are good coins. The single coin outside the pan is the culprit coin.

Case b. If there is change in balance condition – you know which of the 3 coin groups is containing odd coin. Clear all other coins from the scene and in third weighing find which of the 3 coins is of odd weight.

phewwww……at last solved….!!!

1. Divide coins into 4 geoups each containing 3 coins.

2. Now weight any 2 groups out of 4……if the balance shifts then the fake coin is in the weighted group odrwz in the odr 2 groups.

3. Now we are left wd 6 coins. Divide these 6 coins in group of 2 containing 3 coins each.

4. Now weigh any group against the set of 3 coins taken form the good coins. If the balance is balanced thn the fake coin is in 2 group odrwz its in the weighted 3 coins. Also it can be adjudged from the weighing that wether the fake coin is lighter or heavier .

5. now divide this group into a set of one each and and weigh any any two coins. If the condition is balanced 3 coin is faked odrwz the fake coin is determined by the possibility of the coin being lighter and heavier which we found during the 4 step of the procedure.

there is another procedure but it is a bit confusing thats y i havent wrote it here.

If anybody still wants to know can contact me.

yup say that procedure also buddy !!!!!!!!!!!!!!!!!!!

In swcond weighing if the balance is balanced, we know that the fake coin is in 3rd lot but we will not know if it is lighter or heavier. So, this will not work.

first place 6 coins on both the sides of the balance

select the heaviest lot from the two

now divide it into 2 groups of 3 coins each and weigh again.

again select the heavier group.

from the three coins selected take any 2 coins and weigh them

we can thus find the heavier coin

if both have same weight, then the 3rd coin is the heaviest

This is not correct.The side which is heavier may either be due to the heavier coin being on this side or lighter one being on the other side. Try with 4 coins on either side and solve assuming first the faulty coin is heavy and then attempt for either heavier or lighter.

that is correct