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Topic: 3 gold balls (Read 1943 times) |
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Bon BON
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3 gold balls
« on: Jan 20th, 2005, 4:28pm » |
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There is a woman running from pirates because she stole 3 gold balls from them. She comes to a bridge, and there is a sign that says the bridge can only hold 150 pounds. each gold ball weighs 25 pounds, and the woman weighs 100 pounds. she gets across the bridge in one trip with all of the gold balls. she does not roll them. How does she get across? 
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Icarus
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Re: 3 gold balls
« Reply #1 on: Jan 20th, 2005, 6:47pm » |
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This too has appeared before, and again I suggest those of us who have seen it let those who have not provide the answer.
I must say that I prefer this version to the previous ones posted, in that the solution might actually work here (with a certain not unreasonable assumption about the 150 lb limit that I will not mention since it gives too much of a hint). Previous versions had the weights set up in such a way that the intended solution would actually result in collapse.
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JocK
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Re: 3 gold balls
« Reply #2 on: Jan 21st, 2005, 2:33pm » |
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on Jan 20th, 2005, 4:28pm, Bon BON wrote:| she does not roll them. How does she get across? |
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I guess she is not allowed to throw them either? 
I am puzzled by Icarus' remarks. Can the bridge perhaps hold more than 150 lb for (very) short instances?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Icarus
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Re: 3 gold balls
« Reply #3 on: Jan 21st, 2005, 8:15pm » |
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If I may be allowed to reply on Bon BON's behalf, assume the river is too wide for her to throw them across.
As for my remark, when the solution is posted, I believe you will understand it.
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John_Gaughan
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Re: 3 gold balls
« Reply #4 on: Jan 21st, 2005, 8:28pm » |
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:: I think the lady juggled while walking across. ::
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JocK
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Re: 3 gold balls
« Reply #5 on: Jan 22nd, 2005, 5:07am » |
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on Jan 21st, 2005, 8:28pm, John_Gaughan wrote:| :: I think the lady juggled while walking across. :: |
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I was thinking of that as well, but although that could temporarily lead to a force on the bridge corresponding to a weight below 150 lb, peaks well above 175 lb will occur.
According to the laws of physics the 'time average acceleration' on each of the gold balls is zero*, and therefore the 'time average weight' of the juggling lady with the three gold balls has to be 175 lb...
* I.e. moments of negative (downward) acceleration of -g and positive (upward) acceleration of +a cancel each other in the time integral.
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| « Last Edit: Jan 22nd, 2005, 5:15am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: 3 gold balls
« Reply #6 on: Jan 22nd, 2005, 9:31am » |
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I doubt many people could juggle 25 pound balls anyway..
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TenaliRaman
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Re: 3 gold balls
« Reply #7 on: Jan 22nd, 2005, 10:57am » |
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Well,
nothing was mentioned abt the height of the bridge above the ground, so assuming that it was not very high and there is solid ground below it, she went under the bridge 
or
If we assume she had a very long rope, she could have tied these balls with the rope. Then she ran to the other side leaving the balls on this side and once she was over the bridge, she pulled the balls to the other side with the help of the rope.
Thinking ..........
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alien
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The sign that says the bridge can only hold 150 pounds is obviously wrong. The bridge can hold more than a 150 pounds.
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puzzlecracker
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Re: 3 gold balls
« Reply #9 on: Jan 22nd, 2005, 1:50pm » |
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so what it the answer? breaking my head here 
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JocK
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Re: 3 gold balls
« Reply #10 on: Jan 22nd, 2005, 3:40pm » |
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on Jan 22nd, 2005, 10:57am, TenaliRaman wrote:| If we assume she had a very long rope, she could have tied these balls with the rope. Then she ran to the other side leaving the balls on this side and once she was over the bridge, she pulled the balls to the other side with the help of the rope. |
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Indeed: this would lead to sliding of the balls over the bridge, a movement that - unlike rolling - is not forbidden by the riddle.
I assume the lady indeed has a rope which - however - is not long enough to cross the whole bridge. I also assume that bridge is build across water and that the rope can straddle the distance between the bridge and the water. The lady attaches the three gold balls to a rope and - whilst walking over the bridge - drags them through the water. Assuming the balls are not massive and displace each at least some 8.4 liters of water, the buoyancy will ensure the lady gets safely on the other side.
Would the gold balls be massive, this is not going to work, but in that case the lady can go over to plan B: lower the balls deeper in the water and drag the balls over the bottom. Of course in that case she needs a longer rope that can reach the bottom of the water. ::
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| « Last Edit: Jan 22nd, 2005, 3:43pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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puzzlecracker
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Re: 3 gold balls
« Reply #11 on: Jan 22nd, 2005, 7:56pm » |
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It is NP problem... go figure!
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Grimbal
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Re: 3 gold balls
« Reply #12 on: Jan 23rd, 2005, 12:58am » |
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I think the official answer is the one John Gaughan gave on 22. Jan 2005 at 05:28 even though it doesn't pass scientific scrutiny.
But while I am at it, I want to propose another solution: use the Tarski-Banach theorem.
The Tarski-Banach theorem states that it is possible to break a ball into five pieces such that it's possible to combine these pieces into two balls equal in size to the original one. No seams are visible after the operation. No cavities are created under the surface. Using it the other way round, you could combine 2 of the balls into a single one.
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towr
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Re: 3 gold balls
« Reply #13 on: Jan 23rd, 2005, 7:01am » |
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The only problem there is that they're gold balls, which means they're physical objects, not mathematical ones. So supposedly we have to adhere to conservation of energy/mass. So making one ball out of the three simply won't work, the energy released would blow up anything in at least a 3 mile radius..
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| « Last Edit: Jan 23rd, 2005, 7:03am by towr » |
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puzzlecracker
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Re: 3 gold balls
« Reply #14 on: Jan 23rd, 2005, 9:08am » |
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Grimbal could you link me to that answer?
thanks
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Icarus
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Re: 3 gold balls
« Reply #15 on: Jan 23rd, 2005, 1:40pm » |
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The traditional answer is indeed the "juggle the balls" approach. The assumption about the bridge limit I refered to was that the limit is a "sustained weight" limit only, and that the bridge can take momentary loads far in excess. Not something I would be willing to bet my life on, but still the only way juggling would actually work. If I were in the lady's place, the pirates would get one of their balls back!  (Well, no, if I were in the lady's place, I would be out of luck, as I weigh more than 150 lbs ).
Of course, a 100 lb gal with the strength and endurance to juggle three 25 lb balls after running some distance carrying them is surely capable of beating the snot out of those pirates when they come, so perhaps her best bet is to simply look for another way across. 
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I wonder: Which is larger
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JocK
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Re: 3 gold balls
« Reply #16 on: Jan 23rd, 2005, 2:14pm » |
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on Jan 23rd, 2005, 1:40pm, Icarus wrote:| The traditional answer is indeed the "juggle the balls" approach. The assumption about the bridge limit I refered to was that the limit is a "sustained weight" limit only, and that the bridge can take momentary loads far in excess. |
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Hmm... disappointing. A bit like someone posting here the riddle: Write down a mathematical expression that equates to [pi] and uses only three digits. After some failed attempts to find an answer it appears that the intended 'answer' is 22/7...
Thanks for preventing any more time getting waisted on this one.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: 3 gold balls
« Reply #17 on: Jan 23rd, 2005, 2:30pm » |
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on Jan 23rd, 2005, 2:14pm, JocK wrote:| Hmm... disappointing. A bit like someone posting here the riddle: Write down a mathematical expression that equates to [pi] and uses only three digits. After some failed attempts to find an answer it appears that the intended 'answer' is 22/7... |
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[pi] with three digits: [sqrt] (6 [sum][supinfty]k=1 k-2)
[pi] with only two digits (though less informative in a way): 2*asin(1)
[pi] with only one digit (dubious) : ln(-1)/i
[pi] with zero digit: [pi]
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| « Last Edit: Jan 23rd, 2005, 2:42pm by towr » |
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Grimbal
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Re: 3 gold balls
« Reply #18 on: Jan 23rd, 2005, 3:42pm » |
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on Jan 23rd, 2005, 9:08am, puzzlecracker wrote:Grimbal could you link me to that answer?
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I have seen that solution in Demon's Forge, an old adventure-type game (before there were graphic cards...). See the walk-through
http://www.the-spoiler.com/ADVENTURE/Mastertronic/demons.forge.1.html
For the Tarski-Banach theorem, there was an article in Scientific American some time ago in an April section. I wondered some time it it was a joke. In the end it was not. Here is where I got the desctiption:
http://www.cut-the-knot.com/do_you_know/banach.shtml
And yes, I know that the T-B solution is as ignorant of physical reality as the other solution. Even assuming you could do the T-B transformation on the spheres, considering atoms as points, you would get a sphere twice as dense, and certainly unstable.
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SWF
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Re: 3 gold balls
« Reply #19 on: Jan 23rd, 2005, 5:02pm » |
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This may have been mentioned in one of the other threads about this riddle, but she could juggle the balls for a while before stepping on the bridge. With enough strenuous exercise and a well controlled diet, she can lose 25 lbs, allowing her to cross the bridge with the balls in one trip.
If a 100 lb person could cross while juggling 3 balls, using similar reasoning one could make a simple steel sphere submarine without need for complex ballast tanks. Depth could be controlled by a juggler inside.
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Icarus
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Re: 3 gold balls
« Reply #20 on: Jan 23rd, 2005, 8:07pm » |
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on Jan 23rd, 2005, 2:14pm, JocK wrote:
The previous versions posted were worse, setting the bridge limit to exactly the weight of the juggler and one ball. The originators never realized that this would prevent the person from crossing while carrying even one ball, because the dynamic loads of her steps would send the bridge over the limit.
There are several riddles like this I have seen where the proposer totally ignores dynamic loads. Here is another common example.
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