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#11 Prime

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Posted 12 October 2008 - 06:43 AM

Spoiler for my guess

That wouldn't work if all 26 stones weighed, say 23 lb each. (They don't have to weigh different per OP).

However, I can lower my bid to $50 (using only 3 whole number weighs).
Spoiler for 3 weighs

Actually, Imran already used same idea, but didn't optimize the solution.

Edited by Prime, 12 October 2008 - 06:50 AM.

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Past prime, actually.


#12 bonanova

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Posted 12 October 2008 - 09:36 AM

However, I can lower my bid to $50 (using only 3 whole number weighs).

Spoiler for 3 weighs

Bingo. ;)

Honorable Mention to Imran. B))
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#13 imran

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Posted 12 October 2008 - 04:33 PM

Kudos to Prime.

So one should not always trust what bonanova writes :D . That 12$ was a deception :o
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#14 Prime

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Posted 12 October 2008 - 09:33 PM

Kudos to Prime.

So one should not always trust what bonanova writes :D . That 12$ was a deception :o

Thanks. So I won the bid and stand to earn $20. However, I don't feel that we have exhausted the potential of this puzzle. For one thing, I'm still looking for a 2-weigh solution. And although, I haven't found one yet, neither have I proved it impossible.

Consider a follow up weighing job:
There are just two rocks, which can weigh a whole number of pounds from 1 to 4. You are asked to use only one reference weight of your choosing to determine the weight of the two rocks.
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Past prime, actually.


#15 Prime

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Posted 13 October 2008 - 05:31 AM

...
Consider a follow up weighing job:
There are just two rocks, which can weigh a whole number of pounds from 1 to 4. You are asked to use only one reference weight of your choosing to determine the weight of the two rocks.

Disregard that. I made a mistake. You need more than 2 rocks to solve that. See my new topic.
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Past prime, actually.





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