Logic Problem

[Dan]

Here is a logic problem that my friend gave me. If you have heard it before, or if you look up the answer on Google please don't post it here. Some of you will say it is impossible, but I figured out the solution after about 20 min of scribbling it out on paper. It is logic, not trickery.

You have 12 steel balls. One of them is either heavier or lighter than the other 11. All you have to measure with is a simple balance that can tell you if one side is heavier or lighter. Using only 3 measurements, determine which ball has a different mass and whether it is heavier or lighter than the others. Also, you are a terrible judge of weight without the scale, so you cant just pick them up and estimate with your hands.

*solved by ericms*

 

[morgs]

It depends...what type of scale is it? Like a cooking scale or one of those scales that reprisents balance.

 

[Cheomesh]

Hm. I had an answer but I didn't realize it could be either heavier or lighter.

Hmmm...

I haven't found a way to do it in 3 measurements. :|

 

[Dan]

like a see-saw balance. You put stuff on each side and the heavier side will drop down and the lighter side will raise up.

 

[Bad^Hat]

12 balls... and the one can be heavier or lighter? *ponders some more*

Do you know whether it's heavier or lighter beforehand, or is figuring that out part of it?

 

[xcellerate]

I took a 'logic' class and this was actually one of our problems we had to do.

I forgot how it went, unfortunately, but I remember it wasn't as hard as it seemed. My favorite problem was the (apparently) classic 3 door problem that does something like this:

You're on a gameshow and there are 3 closed doors and one of them has a prize behind it, and you pick door #3 and the host says he's going to open one of the other two doors that is empty, so he opens door #1. Now he gives you the opportunity to switch your guess from door #3 to door #2, do your odds increase if you switch doors?

 

[Pesmerga]

[1] [2] [3]

1/3 chance to pick the right door

[2] [3]

1/2 chance to pick the right door



so yes.

 

[Cheomesh]

I'd also like to know if we're aware ahead of time if it is heavier or lighter. If we DO know ahead of time, the problem is easy...

 

[hydrometeor]

I figured out the solution! Go me.

I will represent unknown ball masses as X's and normal ball masses as N's. Balls that are uncertain but weigh equal or are less are L, and balls that are uncertain but weigh more or are equal are M. Brackets will be placed around the balls that are not being weighed. For the balls being weighed, each line is a different side of the scale.

Scenario 1:
{XXXX}
XXXX (in scen. 1 this side will weigh either more or less, becoming L's or M's)
XXXX (in scen. 1 this side will weigh either more or less, becoming L's or M's)

{LM} < (These both probably become N's after this test, if not the answer is obvious)
MMMLLL
NNNNNN

{LNNNNNNNNN}
L
L
The final test determines which of the above 3 L's is the ball that weighs less.

Scenario 2:
{XXXX}
XXXX (in scen. 2 this side will weigh either more or less, becoming L's or M's)
XXXX (in scen. 2 this side will weigh either more or less, becoming L's or M's)

{LM} < (These both probably become N's after this test, if not the answer is obvious)
MMMLLL
NNNNNN

{MNNNNNNNNN}
M
M
The final test determines which of the above 3 M's is the ball that weighs more.

Scenario 3:
{XXXX} < In Scenario 3 These X's remain unknown after the first test
XXXX
XXXX

{XNNNNN} (the x inside the brackets is probably found to be normal after this test, otherwise answer is obvious)
XXX
NNN

{MNNNNNNNNN}
M
M

The final test determines which of the above 3 M's is the ball that weighs more.

Scenario 4:
{XXXX} < In Scenario 4 These X's remain unknown after the first test
XXXX
XXXX

{XNNNNN} (the x inside the brackets is probably found to be normal after this test, otherwise answer is obvious)
XXX
NNN

{LNNNNNNNNN}
L
L
The final test determines which of the above 3L's is the ball that weighs more.

 

[DEATHMASTER]

[1] [2] [3]

1/3 chance to pick the right door

[2] [3]

1/2 chance to pick the right door



so yes.

The question was do they increase if you switch, it's 1/2 whether you do or don't.

 

[MiccyNarc]

Thought I had a solution but I guess not. The fact that it can be heavier or lighter really throws it off

 

[PvtRyan]

Is it allowed to weigh multiple balls at a time, so 6 on one end and 6 on the other end? And what constitutes as a weight measurement?

 

[Cheomesh]

Is it allowed to weigh multiple balls at a time, so 6 on one end and 6 on the other end? And what constitutes as a weight measurement?

Yeah that's wherer I started too.

 

[MiccyNarc]

Is it allowed to weigh multiple balls at a time, so 6 on one end and 6 on the other end? And what constitutes as a weight measurement?

That was what I was going for, but that wouldn't work, as the ball can be either heavier or lighter. So the odd one out could be on either half of the scale, regardless of which side is heavier.

 

[Terminator]

Is the oddball the same physical size as the other balls, i.e it has a different density? if it is then you could drop them all at exactly the same time and observe the ball that either hits the ground first or last, giving an indication of which is the oddball and if it is lighter or heavier than the others in one measurement.

Ok, so it would be near impossible to achieve but it is stll in the realms of possibiltiy and therefore a correct logic awnser, presuming the ball is of the same size.

 

[Dan]

I figured out the solution! Go me.

I will represent unknown ball masses as X's and normal ball masses as N's. Balls that are uncertain but weigh equal or are less are L, and balls that are uncertain but weigh more or are equal are M. Brackets will be placed around the balls that are not being weighed. For the balls being weighed, each line is a different side of the scale.

Scenario 1:
{XXXX}
XXXX (in scen. 1 this side will weigh either more or less, becoming L's or M's)
XXXX (in scen. 1 this side will weigh either more or less, becoming L's or M's)

{LM} < (These both probably become N's after this test, if not the answer is obvious)
MMMLLL
NNNNNN

{LNNNNNNNNN}
L
L
The final test determines which of the above 3 L's is the ball that weighs less.

Scenario 2:
{XXXX}
XXXX (in scen. 2 this side will weigh either more or less, becoming L's or M's)
XXXX (in scen. 2 this side will weigh either more or less, becoming L's or M's)

{LM} < (These both probably become N's after this test, if not the answer is obvious)
MMMLLL
NNNNNN

{MNNNNNNNNN}
M
M
The final test determines which of the above 3 M's is the ball that weighs more.

Scenario 3:
{XXXX} < In Scenario 3 These X's remain unknown after the first test
XXXX
XXXX

{XNNNNN} (the x inside the brackets is probably found to be normal after this test, otherwise answer is obvious)
XXX
NNN

{MNNNNNNNNN}
M
M

The final test determines which of the above 3 M's is the ball that weighs more.

Scenario 4:
{XXXX} < In Scenario 4 These X's remain unknown after the first test
XXXX
XXXX

{XNNNNN} (the x inside the brackets is probably found to be normal after this test, otherwise answer is obvious)
XXX
NNN

{LNNNNNNNNN}
L
L
The final test determines which of the above 3L's is the ball that weighs more.

That is a pretty good solution in that it solves many parts of the problem. But you still have one flaw. In the 2nd step you are weighting 6 balls against 6 N balls which you know are normal. But you only have 4 N balls left from the first test. Your solution requires that you have 2 extra balls of normal mass. Where did you get those extra 2 N balls from?

 

[Dan]

Is the oddball the same physical size as the other balls, i.e it has a different density? if it is then you could drop them all at exactly the same time and observe the ball that either hits the ground first or last, giving an indication of which is the oddball and if it is lighter or heavier than the others in one measurement.

Ok, so it would be near impossible to achieve but it is stll in the realms of possibiltiy and therefore a correct logic awnser, presuming the ball is of the same size.

Your senses are all very dull and inaccurate. You wouldn't be able to tell which ball hit last, you can only accurately read the scale in front of you.

 

[theotherguy]

Is the oddball the same physical size as the other balls, i.e it has a different density? if it is then you could drop them all at exactly the same time and observe the ball that either hits the ground first or last, giving an indication of which is the oddball and if it is lighter or heavier than the others in one measurement.

Ok, so it would be near impossible to achieve but it is stll in the realms of possibiltiy and therefore a correct logic awnser, presuming the ball is of the same size.

Galileo is crying right now.

All objects accelerate at the same rate. It doesn't matter what the mass of the ball is, they will all hit at exactly the same time.

 

[Pesmerga]

The question was do they increase if you switch, it's 1/2 whether you do or don't.

Errr wrong

The intuition of most students tells them that each of the doors, the chosen door and the unchosen door, are equally likely to contain the prize so that there is a 50-50 chance of winning with either selection. This, however, is not the case. The probability of winning by using the switching technique is 2/3 while the odds of winning by not switching is 1/3. The easiest way to explain this to students is as follows. The probability of picking the wrong door in the initial stage of the game is 2/3. If the contestant picks the wrong door initially, the host must reveal the remaining empty door in the second stage of the game. Thus, if the contestant switches after picking the wrong door initially, the contestant will win the prize. The probability of winning by switching then reduces to the probability of picking the wrong door in the initial stage which is clearly 2/3.

 

[Terminator]

Galileo is crying right now.

All objects accelerate at the same rate. It doesn't matter what the mass of the ball is, they will all hit at exactly the same time.

If you were doing the experiment in a vacuum, yes.

That's why i asked if the ball was of the same size and thus more or less dense than the others.
A less dense ball would have less force but the same air resistance and therefore accelerate slower.

 

[Farrowlesparrow]

Errr wrong

Doesn't that rely on the host revealing that you have won, if you initially choose the right door?

 

[Dan]

Doesn't that rely on the host revealing that you have won, if you initially choose the right door?

no...

 

[Farrowlesparrow]

I think I just interpreted differently the part where it says it is apparently explained to students. "If the contestant picks the wrong door initially, then the host must reveal the remaining empty door in the second part of the game" That led me to believe that if you chose the right door, he would simply tell you it was correct.

 

[Dan]

I think I just interpreted differently the part where it says it is apparently explained to students. "If the contestant picks the wrong door initially, then the host must reveal the remaining empty door in the second part of the game" That led me to believe that if you chose the right door, he would simply tell you it was correct.

but he wouldn't

 

[Pesmerga]

No, the host doesn't say if you've chosen the right door or not.

 

[Dan]

The original problem is still unsolved.

 

[Stigmata]

You guys, Farrow was explaining why he was wrong, stop telling him he's wrong now

 

[theotherguy]

If you were doing the experiment in a vacuum, yes.

That's why i asked if the ball was of the same size and thus more or less dense than the others.
A less dense ball would have less force but the same air resistance and therefore accelerate slower.

No. Wind resistance is a function of velocity and a constant of air friction. The force of gravity has nothing to do with it. Two balls of equal size and shape will accelerate at exactly the same rate, meaning their velocities will increase at exactly the same rate, meaning their air resistances will be exactly the same.

 

[mindless_moder]

i think i have it , by my method you will know by the second measurement whether its heavy or light and by the third measurement you will know which ball it is am i on the right track ?.

 

[Terminator]

No. Wind resistance is a function of velocity and a constant of air friction. The force of gravity has nothing to do with it. Two balls of equal size and shape will accelerate at exactly the same rate, meaning their velocities will increase at exactly the same rate, meaning their air resistances will be exactly the same.

No.

What you are trying to tell me there is that a 12" dia foam ball will fall the same speed as a 12" cannon ball in atomosphere, which it clearly dosn't
They both have very different terminal velocities.

 

[Stigmata]

You guys are getting way off track. This problem has nothing to do with air velocities.

 

[Geogaddi]

If you drop a cat off the Eiffel Tower...

 

[Dan]

No. Wind resistance is a function of velocity and a constant of air friction. The force of gravity has nothing to do with it. Two balls of equal size and shape will accelerate at exactly the same rate, meaning their velocities will increase at exactly the same rate, meaning their air resistances will be exactly the same.

You are wrong. The acceleration due to gravity (in a vacuum) is constant, the force isn't. Air resistance is a linear function of drag coefficient, velocity squared, air density, and planform area, but force of gravity is a function of mass, which depends on diameter cubed and object density. So if the object density decreases, the air resistance remains the same, but the force of gravity decreases. If the object density remains the same but the diameter decreases, the force of gravity will decrease more than the wind resistance will decrease. Like Short Recoil said, a plastic ball and a steel ball the same size will not fall at the same speed.

 

[Godron]

No.

What you are trying to tell me there is that a 12" dia foam ball will fall the same speed as a 12" cannon ball in atomosphere.

Yes. Terminal velocity depends on shape and the density of the medium, not weight.

A ball half the weight of another ball would have half the amount of force acting on it, but it also would have half as much mass.
Acceleration = force X mass so...

 

[ericms]

This took me a while, but I am pretty damn sure I'm right. I'm not sure how I will organize myself here...

At first I divided them into two groups of 6, but that got me no where so I did 3 groups of 4 (which I almost had), then 4 groups of 3, then back to 3 groups of 4 which I finally was able to solve after eating. So this is how I did it:

First I consider groups A, B, and C which each have 4 balls in them. So let's say I weigh groups A and B (this is an arbitrary decision as later I can always switch each group out for any other).
So the two main cases of doing this are

1.) Both A and B are the same.
2.) They're of different weight.

Now I'll consider the easier case first which is one.

--------------------
1.1) Since both A and B are the same then I know C is the group with the odd one out. Let's take 1 ball out from B (another arbitrary decision in which I could've chosen A) and another ball out of C. We will label these new groups with 3 balls each B' and
C'. Now I will weigh these two new groups B' and C' (So I have one more chance to use the scale after this).

1.11) This is a subcase in which both B' and C' are equal. Now since they're equal I know that it is the ball from the original group C that I removed that is the odd one out.

Now that I know which ball it is I can weigh this ball with any other one to figure out whether it is heavier or lighter than the rest.

1.12) This is the other subcase in which B' and C' are different. I know the ball is in C' and I also know whether or not this ball is heavier or lighter from weighing A and B previously (I take that weighing plus this weighing and this information is enough to determine if it is heavier or lighter). So what I'll do to figure out which one it is from C' is to take any 2 balls from C' and weigh them.

If these two balls are the same then I know the one I left out from this weighing is the odd one plus I know whether it is heavier or lighter. If they're different then I know which one it is because I know from the previous weighing whether the ball is heavier or lighter and from this I can make the decision.
--------------------

Alright, that finishes everything from case 1 I believe. Now on to the more difficult case.

--------------------
2.1) So A and B weigh different amounts. Let's say that A is lower then B (I'm NOT assuming anything about A or B being heavier or lighter from doing this so this should be fine). What I'll do is take any 2 balls from A plus 1 from B and label this new group A'. I'll then take any 2 of the 3 balls left in B plus 1 of any of the two balls left in A and label this other new group B'. So I have 2 balls NOT A PART of these 2 new groups since I took 6 balls from A' and B' altogether. Now I will weigh A' and B' and this can result in two cases. I'll have 1 more chance to use the scale after this.

2.11) This is a subcase from the second weighing above in which both A' and B' weigh the same. Now since they weigh the same that tells us that one of the 2 balls we excluded from A' and B' is the odd one. Let's take one of the 2 balls and weigh it with any other ball except the other one we excluded (so that we're sure we are weighing it against a normal ball). The 3rd weighing can go down in any of the 2 ways below.

First case is that both of these balls (a normal one and one from the group of 2 that we excluded from A' and B') weigh the same. This tells us that the the other ball we excluded is the odd one out and since we know whether this ball came from A' or B' and furthermore from A or B we know whether it is heavier or lighter then the others (since we made the arbitrary decision above that A was lower on the scale than B).

Second case is that they're both different. This tells us that the ball we picked to weigh against a normal one was the odd one out and by the same logic above we know whether it is heavier or lighter.

2.12) This is the subcase of the second weighing above in which A' and B' are of different weight. Let's make another arbitrary decision that A' is below B'. From this information plus the other arbitrary decision above that A was lower than B we can say that either 1 of 2 balls (JUST ONE BALL!) in A' IS heavier (since 2 of them are from A originally) or that 1 from B' is lighter (since B' has one from group A).

Now what we can do is weigh the two from A' that we suspect might be heavier. If these two are equal then we know that the one in B' is the odd one out and we also know that it is lighter than the rest. If these two are not equal then we, of course, take the heavier of the two.
--------------------

I hope this is right. It took almost as long to write up as it did to solve the damn thing.

 

[Dan]

This took me a while, but I am pretty damn sure I'm right. I'm not sure how I will organize myself here...

At first I divided them into two groups of 6, but that got me no where so I did 3 groups of 4 (which I almost had), then 4 groups of 3, then back to 3 groups of 4 which I finally was able to solve after eating. So this is how I did it:

First I consider groups A, B, and C which each have 4 balls in them. So let's say I weigh groups A and B (this is an arbitrary decision as later I can always switch each group out for any other).
So the two main cases of doing this are

1.) Both A and B are the same.
2.) They're of different weight.

Now I'll consider the easier case first which is one.

--------------------
1.1) Since both A and B are the same then I know C is the group with the odd one out. Let's take 1 ball out from B (another arbitrary decision in which I could've chosen A) and another ball out of C. We will label these new groups with 3 balls each B' and
C'. Now I will weigh these two new groups B' and C' (So I have one more chance to use the scale after this).

1.11) This is a subcase in which both B' and C' are equal. Now since they're equal I know that it is the ball from the original group C that I removed that is the odd one out.

Now that I know which ball it is I can weigh this ball with any other one to figure out whether it is heavier or lighter than the rest.

1.12) This is the other subcase in which B' and C' are different. I know the ball is in C' and I also know whether or not this ball is heavier or lighter from weighing A and B previously (I take that weighing plus this weighing and this information is enough to determine if it is heavier or lighter). So what I'll do to figure out which one it is from C' is to take any 2 balls from C' and weigh them.

If these two balls are the same then I know the one I left out from this weighing is the odd one plus I know whether it is heavier or lighter. If they're different then I know which one it is because I know from the previous weighing whether the ball is heavier or lighter and from this I can make the decision.
--------------------

Alright, that finishes everything from case 1 I believe. Now on to the more difficult case.

--------------------
2.1) So A and B weigh different amounts. Let's say that A is lower then B (I'm NOT assuming anything about A or B being heavier or lighter from doing this so this should be fine). What I'll do is take any 2 balls from A plus 1 from B and label this new group A'. I'll then take any 2 of the 3 balls left in B plus 1 of any of the two balls left in A and label this other new group B'. So I have 2 balls NOT A PART of these 2 new groups since I took 6 balls for A' and B'. Now I will weigh A' and B' and this can result in two cases. I'll have 1 more chance to use the scale after this.

2.11) This is a subcase from the second weighing above in which both A' and B' weigh the same. Now since they weigh the same that tells us that one of the 2 balls we excluded from A' and B' is the odd one. Let's take one of the 2 balls and weigh it with any other ball except the other one we excluded (so that we're sure we are weighing it against a normal ball). The 3rd weighing can go down in any of the 2 ways below.

First case is that both of these balls (a normal one and one from the group of 2 that we excluded from A' and B') weigh the same. This tells us that the the other ball we excluded is the odd one out and since we know whether this ball came from A' or B' and furthermore from A or B we know whether it is heavier or lighter then the others (since we made the arbitrary decision above that A was lower on the scale than B).

Second case is that they're both different. This tells us that the ball we picked to weigh against a normal one was the odd one out and by the same logic above we know whether it is heavier or lighter.

2.12) This is the subcase of the second weighing above in which A' and B' are of different weight. Let's make another arbitrary decision that A' is below B'. From this information plus the other arbitrary decision above that A was lower than B we can say that either 1 of 2 balls (JUST ONE BALL!) in A' IS heavier (since 2 of them are from A originally) or that 1 from B' is lighter (since B' has one from group A).

Now what we can do is weigh the two from A' that we suspect might be heavier. If these two are equal then we know that the one in B' is the odd one out and we also know that it is lighter than the rest. If these two are not equal then we, of course, take the heavier of the two.
--------------------

I hope this is right. It took almost as long to write up as it did to solve the damn thing.

2.12 has flaws in it, you are over-defining A' as the group that is lower than B' as well as the group which has two from A and one from B.

Also, for consistency, it would be a lot easier to read if you normalized your use of "above"/"higher" and "below"/"lower" vs "heavy" and "light". Above could mean higher weight, or it could mean higher position on the scale (lower weight).

If you get it right I will show you my solution. There is definitely more than 1 way to go about it, but they are all kind of similar.

 

[Dan]

Yes. Terminal velocity depends on shape and the density of the medium, not weight.

A ball half the weight of another ball would have half the amount of force acting on it, but it also would have half as much mass.
Acceleration = force X mass so...

your equations don't make any sense in relationship to terminal velocity.

You are only confusing yourself Godron. Terminal Velocity depends on the density of air, the drag coefficient of the object, the mass of the object, and the planform area of the object. lighter objects (all else being constant) have a lower terminal velocity. Try a thought experiment: a giant exercise ball filled with air will fall slower than a giant iron ball of the same size and shape.

 

[ericms]

Well above/below is different from heavy/light the way I use them. Above/below is a generalization of heavy/light since it doesn't assume as much about weight. Also I didn't read what I had written, but I'll start to fix things.

How does 2.12 have flaws in it? I've confused myself now actually.. Hmm. Don't tell me though..I'll figure it out.

It seems that it does have flaws, but in the way that I'm choosing what elements are heavier and lighter.

 

[Dan]

Well above/below is different from heavy/light the way I use them. Above/below is a generalization of heavy/light since it doesn't assume as much about weight. Also I didn't read what I had written, but I'll start to fix things.

How does 2.12 have flaws in it? I've confused myself now actually.. Hmm.

I see what you are trying to say. But if you have a group of balls being weighed, one side is heavier and one side is lighter. If you say one side was above and one side was below, I do not know which is which. Heavy and light and above and below are arbitrary labels which is fine, but it gets confusing when you start interchanging them in the final answer. "X was below, therefore Y is light" does not make sense if the relationship between "below" and "light" isn't clear.

 

[ericms]

Right, sorry about that. I'll clean up the terminology once I figure out 2.12. I was sure I had it, but now the way in which I'm choosing the balls in the last step doesen't make sense anymore.

Edit: How about I break 2.12 into two cases.

First I have that A' is above B' on the scale. Now that means that the element from A' that was originally in B is lighter then the rest OR that ONE of the two elements in B' that were originally in B is heavier. Having the element in A' be lighter is the only one of these two that makes sense since my arbitrary assumption that A be below B would be contradicted if the one of the 2 elements of B' was the odd one.

The second case is that A' is below B'. So this breaks down into the same thing above except I pick one of the two in B' since otherwise the conclusion would contradict my arbitrary decision. Now I can weigh these two and choose the heavier one.