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14 replies to this topic

### #1 bonanova

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Posted 11 October 2008 - 08:10 PM

You learn of a weighing job that's up for bids.
You're out to make some money, so you look into it.

Some guy has 26 rocks, weighing from 1 to 26 pounds, inclusive, not necessarily distinct.
The stones each weigh a whole number of pounds, and the job is to label all the stones with their correct weight.
You have a balance - you know the kind - that gives one of three outcomes for the Left and Right pans:
• L<R
• L=R
• L>R
Luckily, you have an infinite supply of labels marked 1 to 26; what you don't have is any known weights.
You call the warehouse and find that whole-numbered weights cost \$10 apiece and arbitrary-numbered weights cost \$12 each.
Given that you want the job if you can clear \$20 profit, what should you bid?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

### #2 Chuck Rampart

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Posted 11 October 2008 - 10:31 PM

You learn of a weighing job that's up for bids.
You're out to make some money, so you look into it.

Some guy has 26 rocks, weighing from 1 to 26 pounds, inclusive, not necessarily distinct.
The stones each weigh a whole number of pounds, and the job is to label all the stones with their correct weight.
You have a balance - you know the kind - that gives one of three outcomes for the Left and Right pans:

• L<R
• L=R
• L>R
Luckily, you have an infinite supply of labels marked 1 to 26; what you don't have is any known weights.
You call the warehouse and find that whole-numbered weights cost \$10 apiece and arbitrary-numbered weights cost \$12 each.
Given that you want the job if you can clear \$20 profit, what should you bid?

Are the weights of the labels uniform?
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### #3 Ujank

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Posted 11 October 2008 - 10:36 PM

You learn of a weighing job that's up for bids.
You're out to make some money, so you look into it.

Some guy has 26 rocks, weighing from 1 to 26 pounds, inclusive, not necessarily distinct.
The stones each weigh a whole number of pounds, and the job is to label all the stones with their correct weight.
You have a balance - you know the kind - that gives one of three outcomes for the Left and Right pans:

• L<R
• L=R
• L>R
Luckily, you have an infinite supply of labels marked 1 to 26; what you don't have is any known weights.
You call the warehouse and find that whole-numbered weights cost \$10 apiece and arbitrary-numbered weights cost \$12 each.
Given that you want the job if you can clear \$20 profit, what should you bid?

What do you mean by "what should you bid?"
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### #4 SkirlGirl

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Posted 11 October 2008 - 11:19 PM

Spoiler for The lowest bid is...

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### #5 imran

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Posted 12 October 2008 - 12:24 AM

Well I have found one solution. What I read from the puzzle is that a weight of let say 10 pounds or 20 pounds, both cost 10\$ each. And if I have to use let say 10 such weights I need to bid 120\$.

Spoiler for my solution then

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### #6 Masterminded

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Posted 12 October 2008 - 12:29 AM

i did not get this at all
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### #7 Prime

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Posted 12 October 2008 - 02:40 AM

The merchant breaking weight into 4 pieces problem by Bachet de Meziriac, published in 1624, shows how you can weigh the whole number of pounds between 1 and 40 with just ...
Spoiler for Similar problem

So it looks like a high bid, but since no one offered better -- \$60, buying 4 whole numbered weighs.

(I didn't quite get what arbitrary numbered weighs are. Fractional?)
• 0

Past prime, actually.

### #8 Chuck Rampart

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Posted 12 October 2008 - 03:19 AM

The merchant breaking weight into 4 pieces problem by Bachet de Meziriac, published in 1624, shows how you can weigh the whole number of pounds between 1 and 40 with just ...

Spoiler for Similar problem

So it looks like a high bid, but since no one offered better -- \$60, buying 4 whole numbered weighs.

(I didn't quite get what arbitrary numbered weighs are. Fractional?)

I assumed arbitrary meant any non-integer real number. I'm not sure how that would be useful, though.

One more question - do you have to buy all the weights at once, or can you decide which next one to buy based on the results of one weighing?
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### #9 Prime

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Posted 12 October 2008 - 03:39 AM

I assumed arbitrary meant any non-integer real number. I'm not sure how that would be useful, though.

One more question - do you have to buy all the weights at once, or can you decide which next one to buy based on the results of one weighing?

I thought of that. However, since you bid before taking the contract, I assume, you must count on the worst case scenario.
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Past prime, actually.

### #10 Elessar

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Posted 12 October 2008 - 04:01 AM

Spoiler for my guess

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