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## Question

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:

1. L<R
2. L=R
3. 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?

## 14 answers to this question

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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:

1. L<R
2. L=R
3. 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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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:

1. L<R
2. L=R
3. 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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\$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.

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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\$.

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.

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i did not get this at all

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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 ...

... 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?)

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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 ...

... 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?

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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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\$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

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\$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).

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.

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

Edited by Prime

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

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Kudos to Prime.

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

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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.

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...

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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