14 replies to this topic

[bonanova]

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?

[Chuck Rampart]

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?

[Ujank]

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?"

[SkirlGirl]

Spoiler for The lowest bid is...

$0!

If you have 26 stones, each weighing a whole number and knowing at least 2 of the weights (1 and 26 lbs), weigh all the stones and determine which ones are these two extremes. Then, use those two stones as 'weights' for the remaining 24 stones. Just mix-n-match unknown stones against these known stones and voila! No need to buy them fancy, new-fangled weights.

[imran]

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

I can bid 60$.
For this I need weights of 2,6,10,20 pounds.
with a weight of 2 pounds we can identify rocks of 1 pound and 2 pounds.

then we need to put 6 pounds on one side and 2 on other and then we can distinguish rocks of 3 pounds and 4 pounds.

Then by using 6 pounds on one side we can distinguish rocks of 5 and 6 pounds.

and so on.

I am thinking of other solution as I have not utilised any 12$ weight and any label , which make me think that there should be a better answer.

[Masterminded]

i did not get this at all

[Prime]

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

... with just 4 weighs, which are powers of 3:

1, 3, 9, 27. Eg. to weigh 5 pounds put 9 on one side and 1 with 3 -- on the other.
The trick here is, each next weigh is 1 larger than the sum of all smaller ones.
There must be some additional/other trick in this problem, since the stones weigh up to 26 pounds only.

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?)

[Chuck Rampart]

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

... with just 4 weighs, which are powers of 3:

1, 3, 9, 27. Eg. to weigh 5 pounds put 9 on one side and 1 with 3 -- on the other.
The trick here is, each next weigh is 1 larger than the sum of all smaller ones.
There must be some additional/other trick in this problem, since the stones weigh up to 26 pounds only.

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?

[Prime]

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.

[Elessar]

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