A little Problem...
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No, wait:
You can do:
3 and 3, leaving 2 out
Then in the next step you'll only need to do a
1 and 1 to find it, no matter in which group it was. -
2 weighings is the minimum. Unless you know their weights, then the minimum would be one I suppose.
Am I right?
Chris
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wait, are you talking something like this:
or something like this?
but yeah, if i read the wording of the question in a certain way then i'm with chris..
two weighings is the minimum.. or, one is the minimum if we're talking about the first type of scale i posted..(but you can't always do it in these amounts) -
@unknownuser said:
one is the minimum if we're talking about the first type of scale i posted..(but you can't always do it in these amounts)
I'm stumped. How can you do it in one weighing? Sorry, I'm a PC and I can't figure out how.
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One - if you are lucky enough to pick up the heavier one the first try.
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@ecuadorian said:
I'm stumped. How can you do it in one weighing? Sorry, I'm a PC and I can't figure out how.
well, the question says 'what's the minimum amount of weighings?'
so if you have a scale such as the first one i posted, you place one ball in each side.. if one happens to be the heavier ball, you'll know it right away hence 1 weighing.. (and if it's the second type of scale, you need to weigh the two balls separately so it's a minimum of 2 with that scale)
i dunno, there's a few different ways to look at it so remus might have to clarify a bit.
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If you are using the dual scale, not the single scale as the iamges above. In a single scale, if you do not know how much the balls weigh, then 1 weighing doesn't tell you anything. Its the 2nd weighing where you could first find a heavier ball.
But again, it does depend what type of scale we're talking about.
Unless of couse these are bowling balls that are clearly marked how much they wegh, an no weighing is necessary at all
Chris
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None.
You have a scale, so you counter weight using balls as weights.
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@unknownuser said:
so if you have a scale such as the first one i posted, you place one ball in each side.. if one happens to be the heavier ball, you'll know it right away hence 1 weighing.
That would be a total of 4 weighings.
As you place the balls alternating sides, 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. -
@ecuadorian said:
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
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remus, I think you need to see a doctor.
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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?
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Remus never said the problem gave you:
Level 99
Magic 100
Luck 1,000,000For these problems when you say "minumum" it means the minimum without luck and magic.
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@ecuadorian said:
Remus never said the problem gave you:
Level 99
Magic 100
Luck 1,000,000For 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?
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@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.
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Yup. Most people start to do strange calculations with that one.
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7.5/1.5=5
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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.
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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. -
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?)
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