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 Now online (2)big Roger Wilco ...and 121 guests Last 5 registered Oplandisks nothingstar N_loop yipe foxtrotromeo Browse members... |
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You have 10 balls of the same weight. One of them is of DIFFERENT weight. Also you have a balance [pair of scales]. You can use it ONLY 3 times to find that ball. Good luck! |
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You have 10 balls of the same weight. One of them is of DIFFERENT weight. Wait. I thought you said they were all the SAME weight, but one of them is a DIFFERENT weight? YOUR NEW MATHEMATICS FRIGHTEN AND CONFUSE ME!!! :x |
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if you are talking about 10 of the same shape and size but one has a different weight then you: 1) put 5 one side and 5 on the other then discard the 5 balls on the side that is higher on the scale. 2) put 2 and 2 from the remaining 5 on each side and then..... either they are even and the one ball you left off is your winner or..... 3) take the 2 who were lower and the scale, seperate them and which ever is heaver once being weighed for the third time is heaver. |
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ive tried to figure this one out, but ur wrong there creeper, because thats assuming that the odd ball out is lighter than the other balls. if its heavier, youre discarding it right from the get go. |
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actually i was assuming that the odd ball out was heaver. the terminology can get confusing though because i was referring to higher as the height of the scale, not the measure of weight. well either way, do it accordingly to what would make more sense. |
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there is no answer that contradicted it self |
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[little mistake: 9 of same weight except 1 which is of different...] |
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hedphukerr maybe you are wrong phukerrhed ;) |
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There is! Different weight! |
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heavier or lighter? or you can't tell? |
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You have to find out. |
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you seperate them into groups of 3 (1)weigh 2 groups against each other. if they stay the same, you discard those two and work with the remaining group of 3. then (2) take 2 balls out of the remaining group and weight them against each other. if they are the same then the 3rd ball is the winner. if not then (3) exchange one of the balls on the scale for the remaining ball. if the scale does the same thing as last time then the ball left on the scale is the winner, if it is even then the ball taken off is the winner. this is the only way i could think of it using the scale 3 times. the other way would involve the same 3 times but with one step before it all |
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wait now i'm just confused. is it 9 of the same weight and one of a completely different weight? oh who cares. |
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ok it's what my last post said and the only way to explain this is to label each ball which would take way too long |
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Make up two groups of three and one of two out of the eight balls. Weigh the two groups of three against each other. If one is lighter it holds the lighter ball. Take these three balls and weigh one against an other. If one is lighter it is the faulty ball. If they weigh the same the ball left aside is the lighter ball. If when you weigh the two groups of three they weigh the same then the faulty ball is in the group of two. Weigh these two balls against each other and the lighter one is faulty. |
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fuck, submit this to the next myst game |
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"Make up two groups of three and one of two out of the eight balls. Weigh the two groups of three against each other. If one is lighter..." How do you know it? |
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"you seperate them into groups of 3 (1)weigh 2 groups against each other. if they stay the same, you discard those two and work with the remaining group of 3..." And if they don't? |
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