A Puzzle from the Heart of Mathematics?

Question

There are 12 stones. The diamond is embedded in one of the stones. Eleven of the stones weight the same, but the stone containing the jewel weighs either slightly more or slightly less than the others. Find the right stone and find whether it's heavier or lighter. You could use each of the three balance scales exactly once.

Can anyone help me out? My boyfriend and me are working on this one and haven't solved it yet.

You should use each balance scale once. That is, you should put the stones on each balance scale once, and you can't incrementally put stones on one scale -- it would count as many.

Answer 1

NOTE: CHANGE ALL THE "COINS" TO "STONES" AND ALL THE "FAKE" TO "DIAMOND-EMBEDDED".

Name the coins A,B,C,D,E,F,G,H,I,J,K,L

1. Weigh: ABCD||EFGH

2. Weigh: EFGD||IJKH
 If result is the same as "1", as in the right side is heavier/lighter/balance both of the time, then D,H,L either one must be fake. Go 3a.
 If result for "1" is balance and "2" is not, then I,J,K either one must be fake. Go 3b.
 If result for "1" is not balance and "2" is, then A,B,C either one must be fake. Go 3c.
 If result for "1" is both not balance, then E,F,G either one must be fake. Go 3d.

3a.
 If result for "1" & "2" is both balance, Weigh L with another coin.
 If L is heavier, then the fake coin is L and it is heavier.
 If L is lighter, then the fake coin is L and it is lighter.
 If result for "1" & "2" is both unbalance, then balance H with another coin except D.
 If H is heavier, then the fake coin is H and it is heavier
 If H is lighter, then the fake coin is H and it is lighter
 If it is balance, and the result for "1" is heavier||lighter, then D is the fake coin and it is heavier.
 If it is balance, and the result for "1" is lighter||heavier, then D is the fake coin and it is lighter.

3b. Weigh K with J
 If result is J is heavier, and
 result for "2" is lighter||heavier then J is fake and it is heavier.
 result for "2" is heavier||lighter then K is fake and it is lighter.
 If result is K is heavier, and
 Result for "2" is lighter||heavier then K is fake and it is heavier.
 Result for "2" is heavier||lighter then J is fake and it is lighter.
 If result is balance, and
 Result for "2" is heavier||lighter then L is fake and it is lighter.
 Result for "2" is lighter||heavier then L is fake and it is heavier.

3c. Weigh A with B
 If result is A is heavier, and
 result for "1" is heavier||lighter then A is fake and it is heavier.
 result for "1" is lighter||heavier then B is fake and it is lighter.
 If result is B is heavier, and
 result for "1" is heavier||lighter then B is fake and it is heavier.
 result for "1" is lighter||heavier then A is fake and it is lighter.
 If result is balance, and
 result for "1" is lighter||heavier then C is fake and it is lighter.
 result for "1" is heavier||lighter then C is fake and it is heavier.

3d. Weigh E with F
 If result is E is heavier, and
 result for "1" is heavier||lighter then F is fake and it is lighter.
 result for "1" is lighter||heavier then E is fake and it is heavier.
 If result is F is heavier, and
 result for "1" is heavier||lighter then E is fake and it is lighter.
 result for "1" is lighter||heavier then F is fake and it is heavier.
 If result is balance, and
 result for "1" is lighter||heavier then G is fake and it is heavier.
 result for "1" is heavier||lighter then G is fake and it is lighter.


All thanks to my friend.

And Oh yes I just found this answer in murderous maths too. The website is stated below:

Answer 2

put one stone on each side of the balance scale.
If they are the same weight, add one more stone to each side.
again, if the weights are the same, add another to each side, until you put on a pair of stones which cause the scale to tip. You now know that one of these 2 is the stone with the diamond, and all the others don't have a diamond.
On your second scale, place one of the 2 stones that might have the diamond, and another stone which you know doesn't. If they weigh the same, repeat this with the other stone that might have the diamond.
Now you will have found the stone with the diamond.
When you weigh the 2 possible stones against 2 stones you know don't have diamonds, you know which stones might have the diamond, so you would be able to tell if it was heavier or lighters.
If I explained it in a confusing manner I'm sorry. And I don't think the first part means using the first scale more than once, but if it does, sorry again.

Answer 3

Divide the stones into 3 groups of 4, call them A, B and C and then separate each of the groups of 4 into a group of 3 and the remaining 1. We will denote the groups of 3 in A,B,C by a_3, b_3 and c_3 and the singletons by a_1, b_1 and c_1.

1st Weighing: Place A on one scale, B on the other. Note the condition of the scale.

2nd Weighing: Replace A_3 by B_3, and B_3 by C_3 so that the two sides are now B_3, a_1 | | C_3, b_1. Again note the condition of the scale. If it is the same as after the first weighing, then A_3, B_3 and C_3 are all diamondless and we remove them, concluding that a_1, b_1 or c_1 is the diamond stone. If the condition of the scale has changed, then the diamond stone is in A_3, B_3 or C_3. Given the change it should be easy to figure out which of the groups it is and what its relative weight is.

3rd Weighing: Case 1: If it was determined that a_1, b_1 or c_1 is the diamond stone, then set the scales as :

b_1 | | c_1

If this balances, then a_1 is the diamond stone and a previous imbalance tells you its relative weight. If it does not balance, and the previous weighings did balance, then c_1 is the diamond stone. If it does not balance and the previous weighings did not balance, then b_1 is the diamond stone.

3rd Weighing: Case 2: If it was determined that the diamond stone is in A_3 (the cases for B_3 and C_3 are handled similarly), then put one stone from A_3 on the left and one on the right. If it balances, the remaining stone is the diamond stone, if it doesn't you can determine the diamond stone from our knowledge of the relative weights from the first 2 weighings.

Example: Call the stones 1,...,12. Set A_3 = {1,2,3}, a_1 = 4, B_3 = {5,6,7}, b_1 = 8, C_3 = {9, 10, 11}, c_1 = 12. Assume that 2 is the diamond stone and that it is lighter.

1st Weighing: 1,2,3,4 | > | 5,6,7,8

2nd Weighing: 5,6,7,4 | = | 9,10,11,8

We conclude that the diamond stone is one of 1,2,3 and that it must be lighter.

3rd Weighing: 3 | < | 2
We conclude that 2 is the diamond stone.

Example 2: Assume that 8 is the diamond stone and that it is heavier.

1st Weighing: 1,2,3,4 | > | 5,6,7,8

2nd Weighing 5,6,7,4 | > | 9,10,11,8

Since the condition did not change, we conclude that the diamond stone is either 4 or 8 (it obviously can't be 12).

3rd Weighing: 4 | = | 12
We conclude that 8 is the diamond stone and from the second weighing (or the first) we know that it is heavier.

Answer 4

Step 1:
Place 6 stones on each side of the balance. The heavier side will have the diamond in one of its stones.

Step 2:
Take the heavier group, divide it into two groups of 3, and place it on the next scale. Again, the heavier side will have the diamond in one of its stones.

Step 3:
Take the heavier group of 3 stones off the balance, and place 1 stone on each side of the last scale.

If the stones balance out, the remaining stone has the diamond embedded in it.

If the stones do not balance out, then the heavier stone has the diamond embedded in it.

Answer 5

I hope this makes sense (it will be a case argument):

Place 4 stones on each side of the balance:

Case 1) they are equal. Thus it is not any of those stones.
name the remaining stones 1,2,3,4
Place 1 and 2 on the scale (1 on each side).
1.1)they are equal, then it isn't one of them and is either 3 or 4
Place 3 on the scale and one of the stone you know isn't the diamond on the other side:
1.1.1)they balance, then it is 4.
1.1.2)they don't balance, then it is 3.
1.2)1 and 2 do not balance, then it is one of them.
Place 1 on the scale with another stone (not 2):
1.2.1)they balance, then it is 2.
1.2.2) they don't balance, then it is 1.

That was the easy case (I think)
2)The first two sets are not equal. Label the stones on the heavier side a, b, c, and d, and the stones on the lighter side, e, f, g, and h, and the 4 that weren't weighed call them all x.
Place e, f, g, and d and x, x, x, h on the scale:
2.1) They are equal. Thus we know that it is non of them, so e=f=g=d=h=x, and a+b+c+d=a+b+c+x > e+f+g+h=x+x+x+x thus it is one of a, b, or c, and the diamond is heavier.
Place a and b on the balance:
2.1.1)if equal, then it is c
2.1.2) if not equal, then it is the heavier one.
2.2) e,f,g,d>x,x,x,h. Since e,f,g changed sides and the balance stayed the same, we know it's not them. Since it isn't balanced, we know it's not a, b, or c. Thus it is d or h.
Place d on the scale with an x
2.2.1)equal then it is h.
2.2.2)not equal, then it is d.
2.3) e,f,g,d Place e and f on a balance
2.3.1)equal, then it is g.
2.3.2)not equal, then it is the lighter one.

Hope that made sense.

Answer 6

split the 12 stones into 3 groups

**** ****
****

weigh the 2 of the groups on the 1st scale
You'll have 2 possibilities: they either weigh the same or they don't

Let's look at what to do precisely for each of these possibilities

1) if the first 2 groups weight the same
- Throw out the 2 groups you just weighed, they can't have the diamond.
- use your next weighing to do a similar thing with the group you haven't weighed yet (This time one side will be definitely be heavier because you'll have 2 stones on each side of the balance and one side HAS to have the diamond stone), throw out the lighter batch and use your last weighing to solve the puzzle

2) The 1st 2 groups do Not weigh the same
- Throw out the group you haven't weighed and the lighter of the 2 of the groups that you have weighed.
- Follow the same steps as you would have done above.

Answer 7

I will take a stab at this by saying: because 11 of the stones weigh the same...then they could also have the same density

density= weight / volume

Taking an idea from Archimedes first: I would take 3 containers and fill with 100 ml of water apiece( water is 1 gram / milliliter ). Weigh one one of the container and divide the stones into 3 separate but equal numbers (4 stones) weigh each one separately. Two will be identical and the third will be either higher or lower. Do the subtraction and you will get the weight....

I realize this is adding some elements to the puzzle... but in my opinion I could only solve this by thinking out of the box (no pun intended)...

Answer 8

The Heart Of Mathematics Answers

Answer 9

I sat here and mulled over it for half an hour, and I realized that there's no way to find out unless you know whether the stone it's embedded in is denser or lense dense than diamond.

However, lets say you figured out that it's heavier, then you would separate it into 2 stacks of a 6, wiegh them, take the heavier stack and split it into 2 groups of 3, weigh them, take the heavier group, weigh 2 of them, leaving the third out, and if it balances it's the third and if they don't it's the heavier one.
If you find out it's lighter simply invert the logic for lighter stacks.

and Darcy_t2e, it says you can only use the the scales 3 times.

Answer 10

Yes this is a variation of a problem I answered on here about a week ago. It involved finding a counterfeit (lighter or heavier) gold coin among 12 total. You were supposed to do it in three weighings using one balance scale but of course you could use 3 balance scales once. The link to the problem is below. There are 4 answers. Mine is the last. You can always identify the 'odd man out' in 3 weighings, and in all but one case you can tell if it is heavier or lighter as well.

http://answers.yahoo.com/question/index;_ylt=An1aLbLcNLLOZuEVGAboQtbsy6IX?qid=20060629224901AATufsr

Answer 11

Separate the stones into 4 communities of three, communities A,B,C and D. Use the first scale to attempt to stability A and B. If A and B stability the stone is both in team C or D. in the adventure that they don't stability the diamond stone (ds) is on between the perimeters, and one facet will be lighter or heavier. For the sake of debate we are able to assume that A is lighter, and B is heavier. Then weight A with C on the subsequent scale. If A and C at the on the spot are not balanced, all of us do not ignore that the diamond stone is team A, and that that's lighter. If A and C do stability all of us understand the (ds) to be in team B, and that that's heavier, on account that that team become heavier. From the following all of us do not ignore that the stone is both mild or heavy, and we've narrowed it right down to three stones. utilising the finest scale stability 2 of the stones from the gang. If the stability, the third stone is the (ds). in the adventure that they don't, and they are the gang A stones, the lighter is the (ds). in the adventure that they are from team B the heavier is the (ds). Now, enable us assume that communities A and B stability from the start. Now all of us do not ignore that the (ds) is both in team C or D, and we do not understand even if that's heavier or lighter. So, take team C, and weight it antagonistic to team B, which all of us understand to be a conventional team. in the adventure that they stability the stone is in D, in the adventure that they don't the stone is in C and also you may want to tell even if the stone is lighter or heavier by using the way it balances with B. Then seperate the stones from C and weight 2 of them... If C and B stability, you already know the stone is in team D, yet you do not understand its weight. From there you may want to ascertain interior of two which stone become the (ds) and understand even if it become lighter or heavier, yet i do not see a way you may want to understand which of both it become without bending the rules (i.e. taking stones off of the dimensions in my view). I see now from interpreting the above reaction that the answer to thi situation does certainly lie with an set of rules, in simple terms evaluating mixtures by using forking as I proposed gained't remedy the area. Tiger Striped dogs MD