# 12 objects and only three experiments

• LENIN
In summary: Now test U1 vs U3. If they balance then U4 is it if they don't then the balance tells you which one is the odd one.In summary, by separating the 12 objects into 3 groups of 4 and using 3 weighs, it is possible to find the object with the different mass, even without knowing if it is larger or smaller. This method utilizes the balance and compares groups of objects, eliminating those that are equal and narrowing down the search to the remaining objects. With careful grouping and weighing, the odd object can be identified with 100% certainty.

#### LENIN

YOu have 12 objects of the same apperance. 11 of theme have the same mass and one has a different one (larger or smaller). You allso have a ballance but you can use it only three times (and it's the only way to distinguish betwen masses). Is it posibel to find the object with the different mass (be warned that we don't know if the mass is larger or smaller, and that the ballance only showes three stages: balalanced, left side down and right side down). If there is a posibilty explain how it's done and if there isn't explain why it isn't posible.

Yes it is possible.

Separate the 12 objects into 3 groups of 4.
Weigh 2 groups of them.

CASE 1:
If they balance, the 8 objects are good, and the bad object is in the remaining group.
Use 3 of the good objects to weigh against 3 of the remaining ones.
If it balances, the last object that has not been weighted is the bad object.
If it doesn't, one of the 3 suspected objects is either too heavy or too light.
Just compare any 2 of them on the last weigh.

CASE 2:
If it doesn't balance on the 1st weigh, either 1 of the heavier group is too heavy, or 1 of the lighter group is too light. The 4 unused objects are definitely good.
Label the light group of objects L1, L2, L3, L4.
Label the heavy group H1, H2, H3, H4.
The good objects will simply be called G.

2nd Weigh: H1 H2 H3 L1 vs H4 G G G
If it balances, we are left with L2 L3 L4, one of which is too light. Simply compare any 2 on last weigh to find out the bad object.
If left side is heavier, we know for sure that 1 of {H1 H2 H3} is too heavy. Again, compare 2 of them on last weigh.
If right side is heavier, it means either:
L1 is too light or H4 is too heavy. Compare anyone of them with a G to find out.

I hope I am clear. In fact, I think the maximum number of objects we can start with is 13 if we are given 3 weighs.

godchuanz said:
Yes it is possible.

Just compare any 2 of them on the last weigh.

Well if they aren't equal you can't tell if one of them is a lighter or heavier ball compared to the Good balls. But you sort of answered it in the next part.

Again, compare 2 of them on last weigh.
If right side is heavier, it means either:
L1 is too light or H4 is too heavy. Compare anyone of them with a G to find out.
Except that you used 4 weighs when you "compare anyone of them with a G"

Learning Curve said:
Well if they aren't equal you can't tell if one of them is a lighter or heavier ball compared to the Good balls. But you sort of answered it in the next part.
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Except that you used 4 weighs when you "compare anyone of them with a G"

I think you don't get my answer. The explanation is very long if I go into full details I think.
1) when you have 3 balls left, and one of them is too light (or heavy), you can find out which is the one with simply one weigh. Point is, you must already know if the bad ball is too light or too heavy.

2) My "compare anyone of them with a G" is the 3rd weigh. On the 2nd weigh, if the right side is too heavy... then I have either L1 too light or H4 too heavy. Let's say I compare L1 to a G on the 3rd weigh. If it balances, it must mean H4 is too heavy. If it doesn't, then obviously it must be too light.

godchuanz said:
Yes it is possible.

Separate the 12 objects into 3 groups of 4.
Weigh 2 groups of them.

CASE 1:
If they balance, the 8 objects are good, and the bad object is in the remaining group.
Use 3 of the good objects to weigh against 3 of the remaining ones.
If it balances, the last object that has not been weighted is the bad object.
If it doesn't, one of the 3 suspected objects is either too heavy or too light.
Just compare any 2 of them on the last weigh.

CASE 2:
If it doesn't balance on the 1st weigh, either 1 of the heavier group is too heavy, or 1 of the lighter group is too light. The 4 unused objects are definitely good.
Label the light group of objects L1, L2, L3, L4.
Label the heavy group H1, H2, H3, H4.
The good objects will simply be called G.

2nd Weigh: H1 H2 H3 L1 vs H4 G G G
If it balances, we are left with L2 L3 L4, one of which is too light. Simply compare any 2 on last weigh to find out the bad object.
If left side is heavier, we know for sure that 1 of {H1 H2 H3} is too heavy. Again, compare 2 of them on last weigh.
If right side is heavier, it means either:
L1 is too light or H4 is too heavy. Compare anyone of them with a G to find out.[/COLOR]

I hope I am clear. In fact, I think the maximum number of objects we can start with is 13 if we are given 3 weighs.
Correction to case 1)
If you wind up with three untested unknown after two and compare two against each other only if they balance is the 3rd one the odd ball, BUT if the the third test does not balance you only know one might be High or the other could be Low - you'd need one more test!
Instead after one test if you have 8 G's (good) and 4 U's (untested) name them U1 U2 U3 U4. Then test G G vs U1 U2 balance or not brings it down to U1 U2 or U3 U4. Worst case balance and it's U3 U4.
Final 3rd test - G vs U3 still balance means it's U4 but you don't know if high or low. But knowing that was not required.

looks OK but also can reach an answer without knowing if the odd ball is high or low. Last test balances to decide between 3 H's or 3 L's.
Therefore I disagree about 13
- 12 is the max you can deal with in a three test. 13 balls would be to many to work.

I think many of you have problems understanding my solution (case 1). I shall attempt once more...

First, let's establish the fact that if you are left with 3 untested objects, one of which is too heavy, and you have only 1 weigh left, then it is possible to find out which one is the bad ball.
This can be done easily just by comparing any 2 of them, say U1 and U2.
If U1 is heavier than U2, obviously U1 is the bad object, vice versa.
If U1 = U2, then U3 must be the bad object.

That said, you MUST know whether the ball is too heavy/light in order for this to work. This is an important condition.

Solution:
Separate objects into 3 groups of 4. Weigh 2 of the groups.

CASE 1:
The 2 groups balance, thus I am able to identify the "bad group" with the 1st weigh (the untested group).
Now, I have 8 definitely good objects and 4 untested ones (we don't know if it's too heavy/light yet).

On the 2nd weigh, I will compare 3 good objects with 3 untested ones
- if the 3 untested objects are HEAVIER, then surely one of the 3 untested objects is HEAVIER.
- if the 3 untested objects are LIGHTER, then surely one of the 3 untested objects is LIGHTER.

We now know whether it is too HEAVY/light. Therefore, we can now apply the method mentioned above on the last weigh.

(Note: if the 3 untested objects balance the 3 good objects on the 2nd weigh, then they must be good as well and the last untested object must be the bad object. We won't even need the last weigh.)

CASE 2:
Read my first post...

Now, if you think about it using similar methods, 13 is definitely possible. I think that is the maximum you can go.

godchuanz said:
Now, if you think about it using similar methods, 13 is definitely possible. I think that is the maximum you can go.
OK I see what I missed
And since your approch always identifies down to the 12th ball if it is was heavy or light. --
It means if you find even balance all the way though to the Third test you'd know that the bad ball had been replaced with a good ball and was held out as a 13th ball never included in the testing. You'd know the 13th one was the odd one just not if it was high or low.

godchuanz said:
Now, if you think about it using similar methods, 13 is definitely possible. I think that is the maximum you can go.

Since each weighing is either up,down, or balanced, there are only $3^3=27$ possible measurement results. So, with three weighings, if you need figure out whether the ball is light or heavy, then you have to distinguish between twice the number of balls in possibilities, but fourteen balls requires distinguishing between 28 possibilities - that's one to many.

It might be possible to include the possibility that all 13 balls are the same weight.

## 1. What is the significance of having 12 objects and only three experiments?

The number of objects and experiments in a scientific study is determined based on various factors such as the research question, availability of resources, and statistical power. In this particular scenario, it could mean that the study is investigating a small sample size or that the researchers have limited resources to conduct a larger study.

## 2. How can reliable conclusions be drawn from only three experiments?

Although a small number of experiments may seem inadequate, it is possible to draw reliable conclusions if the study is designed and executed properly. This includes controlling variables, randomization, and statistical analysis. Additionally, the results of the study can be replicated by other researchers to ensure the validity of the conclusions.

## 3. Can the results of a study with 12 objects and only three experiments be generalized to a larger population?

It depends on the research question and the study design. If the sample of 12 objects is representative of the larger population and the study is well-designed, then the results can be generalized. However, if the sample is biased or the study design is flawed, the results may not be applicable to the larger population.

## 4. Are there any limitations to conducting a study with only three experiments?

Yes, there are limitations to conducting a study with a small number of experiments. The results may not be as robust and may not account for all possible variables. Additionally, the findings may not be applicable to a larger population. It is important for researchers to acknowledge these limitations and interpret their results accordingly.

## 5. How can the sample size and number of experiments be determined for a scientific study?

The sample size and number of experiments are determined based on various factors such as the research question, resources, and statistical power. Researchers use statistical calculations to determine the minimum sample size needed to detect a significant effect. Additionally, the number of experiments may depend on the complexity of the research question and the amount of variation in the data.