Can You Solve This
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
1 Pick one coin from the first bag, 2 from the second, 3 from the third, 4 from the 4th and so on ... 2 Weight all these (picked) coins together. 3 Subtract the weight from [edit]550 (=10+20+30+40+50+60+70+80+90+100)[/edit] 4 The number you get will tell you which bag has the 9gm coins.:cool: Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 12:39 Monday 24th October, 2005
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
Put 10 coins from one bag, 9 from another, 8 from another, and so on at the same scale. Count all the coins, multiply times 10, take the actual weight, and the number of grams missing is the number of coins you put on the scale from that bag. :cool: -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
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1 Pick one coin from the first bag, 2 from the second, 3 from the third, 4 from the 4th and so on ... 2 Weight all these (picked) coins together. 3 Subtract the weight from [edit]550 (=10+20+30+40+50+60+70+80+90+100)[/edit] 4 The number you get will tell you which bag has the 9gm coins.:cool: Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 12:39 Monday 24th October, 2005
You beat me to it by seconds :mad:!!! :-D -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
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You beat me to it by seconds :mad:!!! :-D -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
Yeah, it's just a case of faster connection. In A.I. they'd say "that's brute force, not real inteligence".:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
Bravo !!! LuisR and Rui A It took me 20 minutes to to solve this so i thot must be worth sharing. :cool: If you have 9 coins with 8 of them 10 gms and One of them 9 gms and a Beam Balance (instead of Weiging machine)
And Two chances to find out which coin is the 9 gm one
much simpler though was related so here you go "Not everything that counts can be counted..." -Albert Einstein
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Bravo !!! LuisR and Rui A It took me 20 minutes to to solve this so i thot must be worth sharing. :cool: If you have 9 coins with 8 of them 10 gms and One of them 9 gms and a Beam Balance (instead of Weiging machine)
And Two chances to find out which coin is the 9 gm one
much simpler though was related so here you go "Not everything that counts can be counted..." -Albert Einstein
Easy. 1 Separate into 3 groups of 3 coins each group. 2 Weight 2 of these groups. If one of them is heavier it has the 9gm coin. If not the other group has it. 3 Sum 2 of the coins of the heavy group. If one is heavier it is the 9gm coin. If not the remaining coin is it.:) Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
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For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
Two. :suss: -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
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Two. :suss: -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
Ok, I accept it, I have no clue!! It's just a wild guess! :-D -- LuisR
Luis Alonso Ramos Intelectix - Chihuahua, Mexico Not much here: My CP Blog!
The amount of sleep the average person needs is five more minutes. -- Vikram A Punathambekar, Aug. 11, 2005
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For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
Huh? Ok, let me try it. So if you have 5 dots to be arranged by 3 to define planes you'd have: combination of 5 by 3= (5!) / ( (5-3)! * 3!) = 10 planes Now, for 1 plane you'd have 0 lines. For each extra plane you add you would have the existing lines plus a number of lines equal to the previously existing number of planes (since the new plane will intercept the existing just once). So:
planes lines
1 0
2 1
3 3
4 6
5 10
6 15
7 21
8 28
9 36
10 45or, as a formula:
lines( 1 )=0 if planes=1
lines( planes )= lines( planes-1)+ planes-1 if planes > 1Makes sense? :confused: Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 13:46 Monday 24th October, 2005
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For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
-
For all the Mathematicians OUT THERE
Five dots are arranged in space so that no more than three at a time can have a single flat surface pass through them. If each group of three dots has a flat surface pass through it and extend an infinite distance in every direction, what is the maximum number of different lines at which these surfaces may intersect one another?
Don't worry, even from here I can clearly hear you say, "Huh?"
:)
"The most beautiful thing we can experience is the mysterious" -Albert Einstein -- modified at 13:25 Monday 24th October, 2005
Well, the whole 5 dots thing is a red herring, since we can only have three dots with planes passing between them. So lets narrow it down to three by setting the extra two dots off in the distance (such that a plane we set down won't pass between them or the other dots). Then we've got three dots, and there are two spaces between them, each allowing one flat surface. So we lay down our flat surfaces and dots like so:
\ /
. \ . / .
\ /
\ /
\/
/\
/ \Then we've got two planes intersecting at one line. (Think of this as a top down view, and the "line" at which they intersect is in the middle of the X, extending through your monitor). Am I right?
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson
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Huh? Ok, let me try it. So if you have 5 dots to be arranged by 3 to define planes you'd have: combination of 5 by 3= (5!) / ( (5-3)! * 3!) = 10 planes Now, for 1 plane you'd have 0 lines. For each extra plane you add you would have the existing lines plus a number of lines equal to the previously existing number of planes (since the new plane will intercept the existing just once). So:
planes lines
1 0
2 1
3 3
4 6
5 10
6 15
7 21
8 28
9 36
10 45or, as a formula:
lines( 1 )=0 if planes=1
lines( planes )= lines( planes-1)+ planes-1 if planes > 1Makes sense? :confused: Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer) -- modified at 13:46 Monday 24th October, 2005
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Well, the whole 5 dots thing is a red herring, since we can only have three dots with planes passing between them. So lets narrow it down to three by setting the extra two dots off in the distance (such that a plane we set down won't pass between them or the other dots). Then we've got three dots, and there are two spaces between them, each allowing one flat surface. So we lay down our flat surfaces and dots like so:
\ /
. \ . / .
\ /
\ /
\/
/\
/ \Then we've got two planes intersecting at one line. (Think of this as a top down view, and the "line" at which they intersect is in the middle of the X, extending through your monitor). Am I right?
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson
-
Well, the whole 5 dots thing is a red herring, since we can only have three dots with planes passing between them. So lets narrow it down to three by setting the extra two dots off in the distance (such that a plane we set down won't pass between them or the other dots). Then we've got three dots, and there are two spaces between them, each allowing one flat surface. So we lay down our flat surfaces and dots like so:
\ /
. \ . / .
\ /
\ /
\/
/\
/ \Then we've got two planes intersecting at one line. (Think of this as a top down view, and the "line" at which they intersect is in the middle of the X, extending through your monitor). Am I right?
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson
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I don't know. It makes sense to me, but I suspect there might be something I didn't notice.:confused::~ Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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10 Bags with 100 coins each Each coin is 10 gms in 9 bags Each coin is 9 gms in one bag (you dont know which ONE bag) You have a weiging machine which gives you the exact weight
You can use the weighing machine EXACTLY ONCE
How will you fnd which bag has coins of 9 gms
You have time till the evening Ask me if any doubt about the question
[DON"T SEE THE ANSWERS BELOW IF YOU SOLVED IT BY YOURSELF DO TELL BE HOW MUCH TIME IT TOOK] "Not everything that counts can be counted..." -Albert Einstein -- modified at 13:03 Monday 24th October, 2005
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What do you think ? :sigh: "Not everything that counts can be counted..." -Albert Einstein
Quartz... wrote:
What do you think ?
I think that I'm right. I don't see a way in which I could position the points and planes in a way that would allow me to get a different answer.
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson