Quiz game #7 The 12 coins

[r_almighty010 872]

 

 

Important :

The quiz is a situation.

The answer to the quiz is the explanation.

So lets start:

 

Situation:

 

There 12 coins and one of them is fake and with a different weight. You have 3 chances to weigh the coins and find out which coin is fake.

 

 

Rules of the quiz:

 

1. I can answer only yes or no.

2. You can only ask questions.

 

The one that finds out how gets 500 crystals.

 

This topic will soon be active

[Weltirex 0]

IN every measuring do you have the same number of coins in every side?

[r_DaSpamer0 14]

First, I moved coin 3 above coin 2 and it touches coins 1 and 2.

 

Second, I moved coin 2 above coin 1 and it touches coins 1 and 3.

 

Third, I moved coin 3 around coin 4 and under coin 1; it touches coins 1 and 4.

 

Fourth, I moved coin 1 under coin 3 and it touches coins 3 and 4.

 

Finally, I moved coin 4 in between coins 2 and 3.

 

The coins are in a straight line.

Was so easy? :P

[r_almighty010 872]

IN every measuring do you have the same number of coins in every side?

yes

First, I moved coin 3 above coin 2 and it touches coins 1 and 2.

 

Second, I moved coin 2 above coin 1 and it touches coins 1 and 3.

 

Third, I moved coin 3 around coin 4 and under coin 1; it touches coins 1 and 4.

 

Fourth, I moved coin 1 under coin 3 and it touches coins 3 and 4.

 

Finally, I moved coin 4 in between coins 2 and 3.

 

The coins are in a straight line.

Was so easy? :P

no

[r_frozen_killer0 0]

they have different values

[r_DaSpamer0 14]

Case 1 - They weigh the same.

That means the counterfeit is among the remaining four. Take two of them, and weigh them one against the other.

If they are equal, take off one and put on one of the remaining 2. If they are still equal the remaining one is fake, if the one you put on is heavier or lighter then that is the fake. (3 weighings)

If they aren't equal, take off one and replace it with a real coin. If the scales don't shift, then the one left on them is fake, if they balance out, the one you removed is a fake. (3 weighings)

 

Case 2 - They are not the same.

That means that the 4 remaining coins are real, and the counterfeit is among those eight. One side of the scales will be higher than the other. We then take 3 coins from the bottom plate and put them aside, take 3 coins from the top plate and put them on the bottom plate, and take 3 real coins (from the 4 we haven't weighed) and put them on the top plate.

If nothing happens, the counterfeit is one of the 2 we didn't move. Weigh one against a real coin - if the same, the other is fake, if different, it is fake. (3 weighings)

If the scales balance out, that means we have removed the real coin, and therefore it is one of the 3 we removed, and heavier than a real one (as they were off the bottom plate). Then, weigh two of the 3 against each other - if one is heavier, that's the fake, if they are the same, the third is the fake. (3 weighings)

If the scales change (ie. the top plate becomes the bottom one), then we have moved the counterfeit coin from the (former) top to the (former) bottom. Therefore it is one of the 3 we moved thus, and it is lighter than a real coin (it was on the top plate). Then, weigh 2 of the 3 against each other. If the same, the third is fake, if one is lighter, it is fake. (3 weighings).

 

EDIT: wait now i just saw that you most answer yes or no... so i cant give the full answer?

[r_frozen_killer0 0]

o.O

[okjes 0]

is the fake coin heavier?

[r_almighty010 872]

Case 1 - They weigh the same.

That means the counterfeit is among the remaining four. Take two of them, and weigh them one against the other.

If they are equal, take off one and put on one of the remaining 2. If they are still equal the remaining one is fake, if the one you put on is heavier or lighter then that is the fake. (3 weighings)

If they aren't equal, take off one and replace it with a real coin. If the scales don't shift, then the one left on them is fake, if they balance out, the one you removed is a fake. (3 weighings)

 

Case 2 - They are not the same.

That means that the 4 remaining coins are real, and the counterfeit is among those eight. One side of the scales will be higher than the other. We then take 3 coins from the bottom plate and put them aside, take 3 coins from the top plate and put them on the bottom plate, and take 3 real coins (from the 4 we haven't weighed) and put them on the top plate.

If nothing happens, the counterfeit is one of the 2 we didn't move. Weigh one against a real coin - if the same, the other is fake, if different, it is fake. (3 weighings)

If the scales balance out, that means we have removed the real coin, and therefore it is one of the 3 we removed, and heavier than a real one (as they were off the bottom plate). Then, weigh two of the 3 against each other - if one is heavier, that's the fake, if they are the same, the third is the fake. (3 weighings)

If the scales change (ie. the top plate becomes the bottom one), then we have moved the counterfeit coin from the (former) top to the (former) bottom. Therefore it is one of the 3 we moved thus, and it is lighter than a real coin (it was on the top plate). Then, weigh 2 of the 3 against each other. If the same, the third is fake, if one is lighter, it is fake. (3 weighings).

 

EDIT: wait now i just saw that you most answer yes or no... so i cant give the full answer?

The first case is correct but the second is not. We only have 3 chances

[r_almighty010 872]

is the fake coin heavier?

We dont know

[r_DaSpamer0 14]

The first case is correct but the second is not. We only have 3 chances

step 1) weigh 1111 (pan X) and 2222 (pan Y)

Case A:

If they are the same then:

- take 3 coins from third group

- take 3 coins from first (or second) group

step 2) weigh 333 and 111

Case A-1:

if they are same

step 3) weigh remaining coin in group 3 with any other coin.

So now you have the wrong coin and whether it is lighter or heavier

Case A-2:

if they are different, you know whether the “black sheep” is heavier or lighter”

step 3) weigh one coin from group 3 on each scale

if they are the same, the culprit is the other coin.

if they are different, you can identify the culprit because you know whether he is lighter or heavier.

 

Case B (after step1):

If they are different, note whether the scales with 1s were lighter or heavier (observation A).

step 2)

weigh 1123(pan X) 1133(pan Y)

if same:

- problem is with one of the three remaining 2s

- step 3) weigh 2 and 2 (trivial with observation A)

if different:

- if pan X shows same result as observation A, then problem is with one of the 1s on pan X. therefore, step 3 is weigh 1 and 1.

- if pan X shows different result from observation A, problem is with the 2 on pan X (or) on the 1s on pan Y. therefore, step 3 is weigh 1 and 1 from pan Y. if same, then problem is with the 2. otherwise use observation A to get the problematic 1.

 

 

Gosh you making me think so bad =/

EDIT: is allowed to use friends? :P

[okjes 0]

I number them first. you first weigh 1234 and 5678.

There are two possibilities. they balance or they don't. If

they balance, then the good coin is in the group 9,10,11,12. So for

the second weighing you put 1,2 in the left and 9,10 on the

right. If these balance then the good coin is either 11 or 12.

 

Weigh coin 1 against 11. If they balance, the good coin is number 12.

If they do not balance, then 11 is the good coin.

 

If 1,2 vs 9,10 do not balance, then the good coin is either 9 or 10.

Again, weigh 1 against 9. If they balance, the good coin is number

10, otherwise it is number 9.

 

 

 

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then

any one of these coins could be that coin. to

proceed, you must keep track of which side is heavy for each of the

following weighings.

 

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against

2,7,8. If they balance, then the good coin is either 3 or 4.

Weigh 4 against 9, a known bad coin. If they balance then the good

coin is 3, otherwise it is 4.

 

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side,

then either 7 or 8 is a good, heavy coin, or 1 is a good, light coin.

 

For the third weighing, weigh 7 against 8. Whichever side is heavy is

the good coin. If they balance, then 1 is the good coin. Should the

weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then

either 5 or 6 is a good heavy coin or 2 is a light good coin. Weigh 5

against 6. The heavier one is the good coin. If they balance, then 2

is a good light coin

 

please a more easy one the next time...

[okjes 0]

please tell me i'm right???

i worked so hard on this... :roll:

[r_AdvocatusDiaboli0 861]

You can answer in yes or no, right? So...

 

hmm...

[okjes 0]

I number them first. you first weigh 1234 and 5678.

There are two possibilities. they balance or they don't. If

they balance, then the good coin is in the group 9,10,11,12. So for

the second weighing you put 1,2 in the left and 9,10 on the

right. If these balance then the good coin is either 11 or 12.

 

Weigh coin 1 against 11. If they balance, the good coin is number 12.

If they do not balance, then 11 is the good coin.

 

If 1,2 vs 9,10 do not balance, then the good coin is either 9 or 10.

Again, weigh 1 against 9. If they balance, the good coin is number

10, otherwise it is number 9.

 

 

 

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then

any one of these coins could be that coin. to

proceed, you must keep track of which side is heavy for each of the

following weighings.

 

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against

2,7,8. If they balance, then the good coin is either 3 or 4.

Weigh 4 against 9, a known bad coin. If they balance then the good

coin is 3, otherwise it is 4.

 

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side,

then either 7 or 8 is a good, heavy coin, or 1 is a good, light coin.

 

For the third weighing, weigh 7 against 8. Whichever side is heavy is

the good coin. If they balance, then 1 is the good coin. Should the

weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then

either 5 or 6 is a good heavy coin or 2 is a light good coin. Weigh 5

against 6. The heavier one is the good coin. If they balance, then 2

is a good light coin

 

please a more easy one the next time...

 

is this right?

[r_DaSpamer0 14]

please tell me i'm right???

i worked so hard on this... :roll:

look on me -.-

btw i posted my answer first.. so if my answer is correct im getting the crystals (cuz i was first :P)

[r_AdvocatusDiaboli0 861]

Uhm... I smell Google. :roll: okjes just replaced a couple of words O:

[hogree 4,179]

Uhm... I smell Google. :roll: okjes just replaced a couple of words O:

 

yeah me too.

[r_DaSpamer0 14]

Uhm... I smell Google. :roll: okjes just replaced a couple of words O:

ROFL

[r_almighty010 872]

The first case is correct but the second is not. We only have 3 chances

step 1) weigh 1111 (pan X) and 2222 (pan Y)

Case A:

If they are the same then:

- take 3 coins from third group

- take 3 coins from first (or second) group

step 2) weigh 333 and 111

Case A-1:

if they are same

step 3) weigh remaining coin in group 3 with any other coin.

So now you have the wrong coin and whether it is lighter or heavier

Case A-2:

if they are different, you know whether the “black sheep” is heavier or lighter”

step 3) weigh one coin from group 3 on each scale

if they are the same, the culprit is the other coin.

if they are different, you can identify the culprit because you know whether he is lighter or heavier.

 

Case B (after step1):

If they are different, note whether the scales with 1s were lighter or heavier (observation A).

step 2)

weigh 1123(pan X) 1133(pan Y)

if same:

- problem is with one of the three remaining 2s

- step 3) weigh 2 and 2 (trivial with observation A)

if different:

- if pan X shows same result as observation A, then problem is with one of the 1s on pan X. therefore, step 3 is weigh 1 and 1.

- if pan X shows different result from observation A, problem is with the 2 on pan X (or) on the 1s on pan Y. therefore, step 3 is weigh 1 and 1 from pan Y. if same, then problem is with the 2. otherwise use observation A to get the problematic 1.

 

 

Gosh you making me think so bad =/

EDIT: is allowed to use friends? :P

Well you said it first and correctly, checkyour crystals in 30 min.