Answer this if ur high IQ.

[CantStopTheMog]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

 

[Node]

Blah blah blah

 

[CantStopTheMog]

Blah blah blah

High Iq answer

 

[registerfasterusing]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

 

[Deleted member 62787]

Quit yapping dickhead

 

[CantStopTheMog]

Blah blah blah

View attachment 4182155

Quit yapping dickhead

This shit is legit high school level at best

 

[JimmyDreamsOfZygos]

6 on either side the first time, 3 from the heavier the second, then put the coins in my ass and jump, the heaviest will fall to the bottom

 

[BigBallsLarry]

This shit is legit high school level at best

easiest question you posted so far. just process of elimination.

 

[Deleted member 62787]

This shit is legit high school level at best

my intelligence is middle school level at best haha get owned idiot pussy fag

 

[CantStopTheMog]

easiest question you posted so far. just process of elimination.

Yes it is easier but it’s cos nobody other than imontheloose can get the answers to the others

Explain the method because yes it is POE but there’s a second step to find if it is heavier or lighter.

 

[Mogsgymmaxx]

I ain’t doing allat

 

[Petsmart]

 

[imontheloose]

Literally make three groups of 4. Weigh two and determine which is heavier or balanced and you have 4 left. Then you can systematically in either case remove one from the group (now you have three) and compare it with another three from before. You can determine this afterwards using common sense.

 

[CantStopTheMog]

I ain’t doing allat

View attachment 4182177

Legit took me under a minute brochacho reading it took longer than solving it

 

[CantStopTheMog]

View attachment 4182181

Blud I am an intern quant I am EMPLOYED as can be

 

[BeanCelll]

4 on each side, and then you make decisions depending on what outcome you get from there?

 

[Mogsgymmaxx]

Legit took me under a minute brochacho reading it took longer than solving it

Blud thinks I have a minute to spare

 

[CantStopTheMog]

Literally make three groups of 4. Weigh two and determine which is heavier or balanced and you have 4 left. Then you can systematically in either case remove one from the group (now you have three) and compare it with another three from before. You can determine this afterwards using common sense.

You only have 3 weigh ins. this would require more than 3 to
A determine which coin specifically has a different weight
B determine whether said coin is heavier or lighter than the others

 

[CantStopTheMog]

4 on each side, and then you make decisions depending on what outcome you get from there?

yes but that’s not a correct answer as u havent given a full method

 

[imontheloose]

You only have 3 weigh ins. this would require more than 3 to
A determine which coin specifically has a different weight
B determine whether said coin is heavier or lighter than the others

No it wouldn’t. First weighing would narrow it to 4. Your second would determine whether the one you removed is odd, and then the third would be a weighing of two coins. It’s always in three weighings.

 

[maximuslaid]

 

[CantStopTheMog]

No it wouldn’t. First weigh would narrow it to 4. Your second would determine whether the one you removed is the counterfeit (so only two needed), and then the third would be a weighing of two coins. It’s always in three weighings.

you don’t know which coin of the 4 assuming it’s not balanced result, is the coin which is lighter or heavier?
If u don’t remove counterfeit on first try u can’t solve it, so this method only has a 25% success rate, therefore incorrect.

 

[CantStopTheMog]

Ok

 

[imontheloose]

you don’t know which coin of the 4 assuming it’s not balanced result, is the coin which is lighter or heavier?
If u don’t remove counterfeit on first try u can’t solve it, so this method only has a 25% success rate, therefore incorrect.

Sir, you do not understand. There’s 24 possibilities, three weighings have 27 outcome patterns, enough to pinpoint one possibility. If the first weighing doesn’t balance, that’s exactly what the second weighing is designed for.

 

[CantStopTheMog]

No it wouldn’t. First weighing would narrow it to 4. Your second would determine whether the one you removed is odd, and then the third would be a weighing of two coins. It’s always in three weighings.

Also this is another Point, you wouldn’t know whether it’s the lighter or heavier pile if theyre unbalanced after the first go as u don’t know whether the counterfeit is lighter or heaviee

 

[CantStopTheMog]

Sir, you do not understand. There’s 24 possibilities, three weighings have 27 outcome patterns, enough to pinpoint one possibility. If the first weighing doesn’t balance, that’s exactly what the second weighing is designed for.

so explain ur method ur using to find the exact coin and whether it is heavier or lighter

 

[Saint Casanova]

No it wouldn’t. First weighing would narrow it to 4. Your second would determine whether the one you removed is odd, and then the third would be a weighing of two coins. It’s always in three weighings.

It’s because you can figure out 27 3^3 possible outcomes but only need 24, 12 times 2 possible outcomes.

Right ?

 

[imontheloose]

Also this is another Point, you wouldn’t know whether it’s the lighter or heavier pile if theyre unbalanced after the first go as u don’t know whether the counterfeit is lighter or heaviee

You don’t need to. Carry both possibilities forward and design the second weighing so three outcomes split those possibilities apart.

 

[imontheloose]

It’s because you can figure out 27 3^3 possible outcomes but only need 24, 12 times 2 possible outcomes.

Right ?

Yes.

Sir, you do not understand. There’s 24 possibilities, three weighings have 27 outcome patterns, enough to pinpoint one possibility. If the first weighing doesn’t balance, that’s exactly what the second weighing is designed for.

 

[Saint Casanova]

Yes.

It’s very easy if you draw one of them tree diagram with 3 sub branches.

We did lots of these problems I’m guessing ?

 

[CantStopTheMog]

It’s because you can figure out 27 3^3 possible outcomes but only need 24, 12 times 2 possible outcomes.

Right ?

The question isn’t about how many outcomes there is, it’s about one of the many methods u can use to get there lmao stop over complicating it.

You don’t need to. Carry both possibilities forward and design the second weighing so three outcomes split those possibilities apart.

 

[BeanCelll]

yes but that’s not a correct answer as u havent given a full method

Oh well manmade numbers and manmade letters hurt my head too much

 

[CantStopTheMog]

It’s very easy if you draw one of them tree diagram with 3 sub branches.

We did lots of these problems I’m guessing ?

not how u work it out because there is more than one variable to consider , there is probability and then also whether it is lighter or heavier

 

[Saint Casanova]

not how u work it out because there is more than one variable to consider , there is probability and then also whether it is lighter or heavier

who says you can’t take into account multiple variables ?

It just makes it easier to group each step.

 

[imontheloose]

so explain ur method ur using to find the exact coin and whether it is heavier or lighter

Treat each as a hypothesis. Coin x is heavy x (hx) or light x (lx). A weighing has three outcomes. Weigh 1, 2, 3, 4 with 5, 6, 7, 8. If equal, odd is among 9, 10, 11, 12. Second weighing, do 9, 10, 11 against 1, 2, 3, (can be any) for example. If equal, the odd is 12. Weigh 12 against 1, heavier/lighter tells type.

If left heavy, one of 9, 10, 11 is heavy. Weigh 9 and 10 which resolves which is heavy. Same logic continues. @Saint Casanova

 

[BaldAutistV2]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

Well you take 6 coins on one side 6 on the other the one that is slightly heavier has the weighted jew coin then you do it again 3v3 coins then 1v1

 

[CantStopTheMog]

who says you can’t take into account multiple variables ?

It just makes it easier to group each step.

How by using a tree diagram will u determine if it’s lighter or heavier?

 

[imontheloose]

It’s very easy if you draw one of them tree diagram with 3 sub branches.

We did lots of these problems I’m guessing ?

It’s perfect for a decision tree, yes.

 

[Saint Casanova]

How by using a tree diagram will u determine if it’s lighter or heavier?

That’s not what I’m saying. I’m saying it just makes it easier to group each step. Obviously you use your reasoning like @imontheloose said.

 

[BaldAutistV2]

compare 6v6

It’s perfect for a decision tree, yes.

He said you can only weigh it 3 times not 11

 

[robotchicken]

I ain’t doing allat

View attachment 4182177

active threat to agepill btw

 

[imontheloose]

compare 6v6

He said you can only weigh it 3 times not 11

You’re stupid. 6v6 does not work. 27>24 hence can be solved in three steps as I have detailed. Do not reply to me again.

 

[CantStopTheMog]

Treat each as a hypothesis. Coin x is heavy x (hx) or light x (lx). A weighing has three outcomes. Weigh 1, 2, 3, 4 with 5, 6, 7, 8. If equal, odd is among 9, 10, 11, 12. Second weighing, do 9, 10, 11 against 1, 2, 3, (can be any) for example. If equal, the odd is 12. Weigh 12 against 1, heavier/lighter tells type.

If left heavy, one of 9, 10, 11 is heavy. Weigh 9 and 10 which resolves which is heavy. Same logic continues. @Saint Casanova

How will u tell if it’s lighter or heavier as u don’t get the weight? you are all forgetting the second component of this question lmao. Ur not just working out which coin has a different weight, you are also working out if it’s lighter than the others or heavier

 

[Mogsgymmaxx]

active threat to agepill btw

Wdym

 

[BeanCelll]

Well you take 6 coins on one side 6 on the other the one that is slightly heavier has the weighted jew coin then you do it again 3v3 coins then 1v1

You need 4v4

 

[arifon]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

probably any 8 yr old can solve this but for the org members it’s tough for sure

 

[CantStopTheMog]

It’s perfect for a decision tree, yes.

It literally isn’t lol

 

[.Cx]

this one is actually super easy tbh

You have 12 visually identical metal coins. One of them is counterfeit — it looks the same but its weight is different (you don’t know whether it’s heavier or lighter). You have a balance scale (compares two groups and tells you which side is heavier, or if they balance). Using the scale only three times, identify which coin is counterfeit and whether it is heavier or lighter.

@Gaygymmaxx
@imontheloose



Rules: you must decide which coins to put on the scale each weighing based only on previous results (i.e., adaptive strategy). Give the plan and how the outcomes uniquely identify the counterfeit coin and its weight difference.

This ones too easy to dignify with a response.

 

[imontheloose]

It literally isn’t lol

Draw 24 leaves.

 

[CantStopTheMog]

probably any 8 yr old can solve this but for the org members it’s tough for sure

these guys r literally not listening they’re doing the easy part like I’d expect but they’re not even answering the second part lmao