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Originally Posted by
Scheherazade
The DA is a woman and the killer's mother?
Yep.
Women who've been thinking about the problem for a long time tend to be very cross with themselves when you tell them the answer.
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Pièce de Résistance
Next puzzle (remember this one from my childhood
)
A jeweller has ten bags of golden coins, each containing ten coins; however, he is informed that all the coins in one of the bags are counterfeits: they weigh one gram less than the regular coins in other bags.
How can he discover the bag with the counterfeits by using the scales only once?
~
"It is not that I am mad; it is only that my head is different from yours.”
~
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Wild is the Wind
Can he carry all the bags and stand on the scales and then throw the bags one by one onto the floor around him? He would know which bag weighed differently from all the others by the irregular weight change. This would only be strictly using the scales once, wouldn't it, as his weight stays on the scales until the end?
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Pièce de Résistance
No, he cannot do that; it is a small one used by jewellers.
And the method you suggest would mean using the scales up to 10 times. He uses the scales only once; just one reading.
~
"It is not that I am mad; it is only that my head is different from yours.”
~
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Originally Posted by
Scheherazade
No, he cannot do that; it is a small scale used by jewellers.
And the method you suggest would mean using the scale up to 10 times. He uses the scales only once; just one reading.
Having used the scales, can he use other handy stuff?
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Registered User
Got it! (Maybe!) But it assumes that we know how much a non-counterfeit coin weighs.
1. Number the bags from 0 to 9.
2. Then, take one coin from bag number one, two coins from bag number two, three from number three, etc. (Don't take any coins from bag number zero.)
3. The total number of coins will be 45 coins.
4. ASSUMING that each non-counterfeit coin weighs some certain amount 'X', the equation "45X - WEIGHT" will yield a number of grams that matches the number of the counterfeit bag.
Scher, do we begin knowing how much a non-counterfeit coin weighs???
If not, then my plan falls apart.
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Originally Posted by
billl
Got it! (Maybe!) But it assumes that we know how much a non-counterfeit coin weighs....
That's very ingenious, even with the assumption - but it also assumes that the scales are like this. I'd assumed that they were like this, and that no marked weights were provided, so all you have to work with is the bags.
Scher?
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Pièce de Résistance

Originally Posted by
billl
Got it! (Maybe!) But it assumes that we know how much a non-counterfeit coin weighs.
1. Number the bags from 0 to 9.
2. Then, take one coin from bag number one, two coins from bag number two, three from number three, etc. (Don't take any coins from bag number zero.)
3. The total number of coins will be 45 coins.
4. ASSUMING that each non-counterfeit coin weighs some certain amount 'X', the equation "45X - WEIGHT" will yield a number of grams that matches the number of the counterfeit bag.
Scher, do we begin knowing how much a non-counterfeit coin weighs???
If not, then my plan falls apart.
Yes, we do know how much each coin weigh: 10g.
(Was expecting someone to ask this question sooner
)
Bill's answer is correct:
- Put labels on the bags, numbering them from 1-9.
- Take the same number of the coins from each bag as they are indicated on their labels.
- There will be 55 coins (not 45), making the expected weight 550g; however, because of the counterfeits, total weight should be less than that and missing grams will indicate the number of the bag the coins come from.

Originally Posted by
MarkBastable
That's very ingenious, even with the assumption - but it also assumes that the scales are like
this. I'd assumed that they were like
this, and that no marked weights were provided, so all you have to work with is the bags.
Scher?
Earlier, when I answered Silas' question, I did mention that it was one of those used by jewellers.
Your turn, Bill!
~
"It is not that I am mad; it is only that my head is different from yours.”
~
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Registered User
HA YES!
but....
bag 1 + bag 9 +
bag 2 + bag 8 +
bag 3 + Bag 7 +
bag 4 + bag 6 +
bag 5
= 45.
Bag zero adds zero. (Imagine if Bag zero had the counterfeits...)
Wow, I feel pretty bad-***. But now I need to think of (remember) a puzzle, and that one is gonna be hard to match.
Last edited by billl; 10-26-2010 at 03:37 AM.
Reason: remember...
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Registered User
Hmm, OK, I hope that I am around enough to answer questions. Or that someone doesn't give the answer right away, if it is too obvious...
This is like the one Mark did earlier, but if you haven't heard it before, you might need to ask some questions in order to work out the answer... (Sort of like the game twenty questions... I will give only yes or no answers!)
HERE IS THE SITUATION:
A man is lying in a field, dead, next to a large rock. His body is a perfect specimen of manly health--no marks, no bruises, no visible injuries. But he is dead. What happened?
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Pièce de Résistance

Originally Posted by
billl
HA YES!
but....
bag 1 + bag 9 +
bag 2 + bag 8 +
bag 3 + Bag 7 +
bag 4 + bag 6 +
bag 5
= 45.
Bag zero adds zero. (Imagine if Bag zero had the counterfeits...)
Yes, I see how this works out but in my solution you take 10 coins from bag no.10 as well, making the total 55.
Going back to your question, is the rock crucial in his death?
~
"It is not that I am mad; it is only that my head is different from yours.”
~
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Registered User
(btw, I can see how a 'bag 10' instead of 'bag zero' changes the math...)
Um, yes, the rock plays a role in his death.
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Convinced, erroneously, of his manifest destiny, he gave himself a heart attack trying to get the sword out of the stone.
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Pièce de Résistance

Originally Posted by
billl
Um, yes, the rock plays a role in his death.
Since there were no visible injuries, is it safe to assume that he was not crushed by this rock?
How large is it? (And, more importantly, does size matter?)
Is it a regular rock by the way? I.e., "a large piece of stone" in the sense we use in everyday conversation?
~
"It is not that I am mad; it is only that my head is different from yours.”
~
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Registered User

Originally Posted by
MarkBastable
Convinced, erroneously, of his manifest destiny, he gave himself a heart attack trying to get the sword out of the stone.
A nice attempt, but no.

Originally Posted by
Scheherazade
Since there were no visible injuries, is it safe to assume that he was not crushed by this rock?
How large is it? (And, more importantly, does size matter?)
Is it a regular rock by the way? I.e., "a large piece of stone" in the sense we use in everyday conversation?
He was not crushed. Let's say that the stone was the size of a loaf of bread. It is not a "regular rock", I guess, but it is a large piece of stone, in the sense that we usually use.
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