A little Problem...

jeff hammond

That would be a total of 4 weighings.
You start weighing 1 vs 1, then 2 vs 2, then 3 vs 3, and then finally 4 vs 4 if you haven't found it yet.

you're talking about the maximum amount of weighings.. the question says minimum

but going that route with the dual scale to find the max amount.. i'd try
3vs3 then mono-y-mono..
you can always find the oddball using a maximum of 2 weighings with the dual scale.

edit- oh wait, that's the same method you describe earlier.. so yeah, i agree with you on that

boofredlay

remus, I think you need to see a doctor.

Ecuadorian

I guess we need another problem now.

If I want to buy seven rabbits and a half, and each rabbit and a half costs one dollar and a half, how much do I have to pay?

Ecuadorian

Remus never said the problem gave you:

Level 99
Magic 100
Luck 1,000,000

For these problems when you say "minumum" it means the minimum without luck and magic.

jeff hammond

@ecuadorian said:

Remus never said the problem gave you:

Level 99
Magic 100
Luck 1,000,000

For these problems when you say "minumum" it means the minimum without luck and magic.

right, i agree and hope that's how it is..
but - there's nothing worse than someone asking you a logic question only to find out it was really some weak play_on_words type of thing..
(well, i guess i can think of worse things but... )

someone has to build up some cred with these types of questions with me or i have to try to eliminate the loop holes first..

so who has another one?

jeff hammond

@ecuadorian said:

I guess we need another problem now.

If I want to buy seven rabbits and a half, and each rabbit and a half costs one dollar and a half, how much do I have to pay?

$7.50

dividing by 1.5 then multiplying by 1.5 cancel each other out so it's a buck per rabbit.

Ecuadorian

Yup. Most people start to do strange calculations with that one.

xrok1

7.5/1.5=5

remus

Well done people, its not a trick question: 2 is the correct answer as ecuadorian said (and the scales are the 1st type you posted jeff.)

edit: doh, put 3 instead of 2. Corrected.

Alan Fraser

I'd have said $10.50, Miguel. You said 7 rabbits and a half...that's 7 whole rabbits plus a half rabbit. Given, that in Ecuador, you apparently sell rabbits in the odd amount of 1.5 rabbits, you'd need to buy 7 portions of them to get 7 whole rabbits...then have a whole lot of half-rabbits left over.
I always was something of a pedant.

Gaieus

I agree with Eric - go and see a doctor, Remus. And when you've done so, also a psychiatrist (since who cares which ball weighs slightly more or less?)

remus

Who cares! what do you mean who cares?! we're addressing fundamental problems in comparative ball analysis and scalar logic

Ecuadorian

Another one:

I have two jars full of a liquid amoebas eat. These amoebas multiply by 2 every three minutes. Jar A starts with two amoebas and after two hours it's full of them. Jar B starts with just one. How much time will it take it to fill?

remus

2 hours 3 minutes, assuming the jars are the same size.

reasoning: after 3 minutes it will have multiplied, so you'll have 2 amoebas and be left with the same starting conditions you had in jar A.

Ecuadorian

Anyone has another one?

Anssi

@remus said:

Well done people, its not a trick question: 3 is the correct answer as ecuadorian said (and the scales are the 1st type you posted jeff.)

No, Remus, two weighings is enough, when you start with the 3+3 method I think Jeff posted first:

First weighing: You select randomly 3+3 balls and weigh them. The next step depends on your result.

Second weighing: If the 6 balls were of equal weight you weigh the 2 balls left, the heavier is one of them. Otherwise, you choose randomly 2 balls out of the heavier set of 3 and weigh them. If they are equal, the heavy one is the one left out.

In real life, figuring this out would probably take more time than to weigh the balls in three steps
Anssi

pilou

About the first problem with this and 12 same aspect objects (one is different weight)

You must say what is the different object and if it's less or more weight than the current!
In how many minimum weigh-in can you made that?

Whithout see result on the Net of course!

jeff hammond

pilou

that would take 3 weighings

6 vs 6 then
3 vs 3 then
1 vs 1

[or 5vs5, 2 vs 2, 1 vs 1 leaving the possibility that you might find it in 2 tries if the 5 vs 5 results are the same]

could also do:
4vs4
2vs2
1vs1
which still needs 3 weighings

jeff hammond

@xrok1 said:

7.5/1.5=5

while that is true, it's not the proper answer to the rabbit question.

a different way to think about that one is to first ignore the total amount of rabbits and figure out the cost first:

a rabbit and a half costs a dollar and a half therefore one rabbit costs one dollar.. so 7.5 rabbits costs $7.5

dale

OK now if you model and render the eight balls and seven rabbits, the price of the rabbits will go up,(because thats going to take a while, and time is money) and the weight of the balls will vary based on the visual weight they are allotted, now things are different! (oh render engine of your choice by the way )