A little Problem...
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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?
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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.
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Anyone has another one? -
@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 -
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!
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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 -
@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
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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
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@ Jeff
Number's 3 answer is good but explanation seems lack of precision and sybilline for external visitor
What about object number 1,2,3,4,5,6,7,8,9,10,11,12?
Does is it less or more heavy or current? -
@unknownuser said:
@ Jeff
Number's 3 answer is good but explanation seems lack of precision and sybilline for external visitor
What about object number 1,2,3,4,5,6,7,8,9,10,11,12?
Does is it less or more heavy or current?haha..
i'm definitely not the best at explaining myself.. if i tried to explain using object numbers, it would get a lot more confusing i thinkbut here goes an example.. let's say the heavier object is #9
place objects 1-6 on one side then objects 7-12 on the other.. 7-12 will weigh more so that eliminates objects 1-6
place 7,8, & 9 on one side then 10, 11, & 12 on the other.. 7-9 will weigh more so that eliminates 10-12..
place 7 on one side and 8 on the other... they weigh the same so that means 9 is the oddball.
[EDIT] and then as Anssi said, in real life, you know you're going to test #9 against some of the other objects so the real world answer is at least 4 if not more
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@unknownuser said:
You can't say first 9 is heavy!
ha
well i did that for sake of explanation..regardless, i can figure out the answer in about 5 seconds but i could sit here all day long trying to explain how
it's a fault of mine.. -
@ Jeff : yes but...no
You can't say first 9 is heavy!
You must make first all combinaisons and then say for this combinaison 9 is heavy !
So draw first all the tree !Here your proposition
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If i take you 11 oranges and one false your explanation don't work
Only with a 1/24 hasard -
really? the way i see it, you can always find the the heavy one in 3 tries.. you mind showing the 1 out of 24 when it wouldn't work?
further, using a different starting technique using 12 objects (5vs5), it's possible to find the oddball in 2 tries.. sometimes (17% chance of doing it 2)
let's say the heavier object is #11..
place 1,2,3,4,5 on one side and 6,7,8,9,10 on the other.. they weigh the same so that eliminates 1-10
place #11 on one side and #12 on the other..
11 weighs more so you've found it in 2 weighings.
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Annsi, sorry, typo. 2 is the right answer to the original question
Ive got another one for you.
(This problem is an extended case of the above problem.)
The aim is to transfer all the discs from one pole to another without placing a larger disc on a smaller disc at any point.
The rules for moving the discs:
-you can only move 1 at a time,
-a move consists of moving one disc from one pole to another without changing the position of any other discs;
-a disc must always be placed on a larger disc or directly on the base.If i can move the discs at a rate of 10 a minute, how long would it take me to complete a puzzle with 64 discs?
For additional cool points, can you prove your answer? and how does adding a 4th pillar effect the solution?
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@remus said:
If i can move the discs at a rate of 10 a minute, how long would it take me to complete a puzzle with 64 discs?
wait, you didn't mention what the finished puzzle should be.. do you mean start with the picture and move the stack to a different pole following your guidelines?
[edit1]oh wait.. 3 poles, all 64 discs are off the poles, then build one stack on one pole?
[edit2] ok, i think i got it now
.. start with a stack of 64 discs on one pole then transfer them to another pole ? -
@Jeff
Of course yes, it's just the methodology
In the real world you will be obliged to make all combinaisons first till find a solution and say this is the false object!
Try it in the real world and you will see that i am right
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@unknownuser said:
wait, you didn't mention what the finished puzzle should be.. do you mean start with the picture and move the stack to a different pole following your guidelines?
Thats it, the aim is to move all the discs to another pole following the rules, so for 2 discs the solution would look like this:
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