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Guest Message by DevFuse

Started by bonanova, Oct 11 2008 08:10 PM

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14 replies to this topic

#11 Prime

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Posted 12 October 2008 - 06:43 AM

Spoiler for my guess
$30
buy 1 2 pound weight. then you can find if there are any 1 or 2 pounders, using those and the one you bought, you can find 3,4,5... etc

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.

0

Past prime, actually.

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#12 bonanova

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
The weighs 2, 6, 18 could determine anything up to 27 lb, as long as a whole number weight is guaranteed.
With 2 lb we can find wheter a stone weighs 1 (less than 2), 2, or more.
Then with 6-2=4 we can sort out 3, 4, or larger.
6 isolates 5, 6, or larger.
6+2=8 helps finding 7, 8, or larger.
Then use 18-6-2=10 for 9, 10, or larger.
And so on.

Bingo.

Honorable Mention to Imran. B))

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

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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

. That 12$ was a deception

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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 . That 12$ was a deception

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.

0

Past prime, actually.

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#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.