Puzzle for the Smart Kids :) - GoFuckYourself.com

Pipecrew · 03-14-2002, 08:14 AM

There are ten seemingly identical stacks, each containing ten coins. One of these stacks, however, in its entirety, is counterfeit and weighs differently from the genuine coins, presumed to be only detectable by employing a scale.

You are challenged to find the absolute minimum number of weighings necessary to indisputably locate that counterfeit stack. Be advised, that any and each weight reading is considered a separate weighing. The weights of the two types of coins may be assumed, whereby you can facilitate simplicity in mathematics.

Your answer should include an explanation of your manipulation of the coins, in order to justify the minimum number of weighings in your solution and encourage discussions among participants.

This puzzle contains no deceptive wording. The correct strategy of solving is purely logical, totally realistic and physically executable.

==========

Some guy posted this on my sites chat board, if you got any idea's post it here and I will post it, shutting him up for good (one of those guys whose life mission, is to prove he is smarter then everyone)

Za Ha · 03-14-2002, 08:22 AM

Too much to read

But answer the question in my signature.... now thats hard.

Living For Today · 03-14-2002, 08:26 AM

I started attempting to answer this but my brain just cant be bothered

Pipecrew · 03-14-2002, 08:55 AM

haha well if anyone gets motivated or can read that much answer the damn thing

Jakke PNG · 03-14-2002, 08:58 AM

Hehe...
just weight each stack once

The one with different weight is wrong.

Hypo · 03-14-2002, 09:01 AM

4 weightings, 5 max. First weight 4 and 4. If they are equal then one of the two remaing stacks is faulty. If they are unequal separate one group of 4 into 2 n 2 and weight - if equal the other 4 is faulty. Separate those into 2 and 2. Weight. Weight. Thats about it. You can pretty much figure out all the options.

Jakke PNG · 03-14-2002, 09:04 AM

My method was better. I'd do it that way.

NetRodent · 03-14-2002, 09:07 AM

I would need 3 to 5 attempts at weighing the stacks of coins to find the counterfeit stack.

1. Pick 1 stack of coins and exclude it. Weigh the remaning 9 stacks of coins.

2. Divide the nine stacks into 3 sets of 3 stacks and weigh two of them. The third stack I deduce the weight of by subtracting the weight of the two known stacks from the measurement in step 1.
If I am lucky all 3 sets of 3 stacks weigh the same meaning the stack I excluded in step one is the counterfeit stack.

3. If not, at least I know which of the sets the counterfeit stack is in. From that set I would pick any stack at random and weigh it. By this point by comparing the weights I measured in step 2 I know what a stack of real coins should weigh. If the weight is not proper for a real stack of coins, I've found it on the 4th weighing. If it is proper, I take one of the two remaining stacks and weigh it. This fifth weighing is the last one I need. If the stack wieghs the correct amount, I know the other stack is counterfeit. If It doesn't weigh the correct amount then it is the counterfeit stack.

Thus 3-5 weighings are all that is needed.

Hypo · 03-14-2002, 09:09 AM

On second thought, I have a better answer - just 1 weighing will do! Take 1 coin from stack 1. Two coins from stack 2. Three coins from stack 3 etc. Weigh all the coins, divide the excess weight with the weight of the counterfeit coin. You get the answer.

EasyDialers · 03-14-2002, 09:09 AM

Quote:

Originally posted by Za Ha
Too much to read

But answer the question in my signature.... now thats hard.

i would have to say no.....or yes? I give up. Just give me the answer

NetRodent · 03-14-2002, 09:13 AM

What sort of device are we able to use for measuring weight? I see that Hypo assumed it was a balance thus being able to compare two sets in one weighing. I assumed it was a calibrated scale allowing only one stack or set of stacks to be weighed at a time.

NetRodent · 03-14-2002, 09:14 AM

Quote:

Originally posted by Hypo
On second thought, I have a better answer - just 1 weighing will do! Take 1 coin from stack 1. Two coins from stack 2. Three coins from stack 3 etc. Weigh all the coins, divide the excess weight with the weight of the counterfeit coin. You get the answer.

Wouldn't this only work if you know the weight of counterfeit coin and a real coin ahead of time?

HeadPimp · 03-14-2002, 09:41 AM

Fuck that! Lets go with 0 weighings. The say that the coins are of different weights but of identical dimensions indicated that they are of different densities. By using a weight with a point on it, dropped from a fixed height, you can create a dimple in each coin. Teh coin that has a different size dent is the fake.

OR!

Arrange the stacks around the rim of a large disk that is perfectly balanced at its center. The stack that either comes out higher or lower is the fake stack.

FADE19 · 03-14-2002, 09:45 AM

1+1=3

ADL Colin · 03-14-2002, 10:31 AM

Do we know if the counterfeit coins are lighter or heavier than normal? Or that has to be deduced?

ADL Colin · 03-14-2002, 10:32 AM

A balancing scale or a regular scale?

ADL Colin · 03-14-2002, 10:38 AM

Here .. since you wanna prove him wrong ...

Fastest way to find a solution is to look it up ;)
http://scientium.com/drmatrix/puzzles/mg1puz5.htm

You might wanna change the wording of the answer.

bauhaus · 03-14-2002, 10:39 AM

I think the answer is 6 at most....

1. weigh 5 stacks record

2. weigh 5 stacks record, then continue with the lightest/heaviest (depending on what you assume the variable to be)

3. now you have 5 stacks left.... 4 separate weights are taken next ex. 1 2 3 4 5

weight of 1+2, 2+3, 3+4, 4+5

4. now you have taken 6 weights.....out of the 4 weights above you can figure out which one is heaviest/lightest (again depending on what you set you known variables to be) using comparitive measures....

Unless it is a lateral thinking trick....it can't be less

ie...you can use a counterbalanced scale (therefore you could balance the scale against the different stacks, thereby reducing the amount of total weights needed)

ADL Colin · 03-14-2002, 10:46 AM

The answer is one.

Very clever problem.

bauhaus · 03-14-2002, 10:53 AM

The weights of the two types of coins may be assumed, whereby you can facilitate simplicity in mathematics.

If this is assumed then you could pick the right one first? that would give you an answer of one? Colin how else?

FADE19 · 03-14-2002, 10:55 AM

CRAP....I was soooo close I had 3....damn it took me awhile to come up with that formula too.

ldinternet · 03-14-2002, 10:57 AM

Colin, the given answer has its limitations in that it assumes that you can choose the mass overage for the counterfeit coins. How about having a standardised overage of mass for the counterfeit coins of, for example, 0.02kg?

bauhaus · 03-14-2002, 10:57 AM

just read it now.....I didn't realize you could disturb the stacks (hence the lateral thinking waiver) should have though, otherwise the addition to the equation of stacks of 10 would have been meaningless.....

bauhaus · 03-14-2002, 11:02 AM

ID I see what you are saying...but in your example of a difference of .02kg....its easy because the the stacks are in groups of 10 and the kilograms are metric...you would have mind fucked us if you used your fuckin imperial weights..

If you were not sure of anything my way would work using the minimun amount of weighs.....if you could not even assume the counterfeits were more or less you you just have to do the 1+2, 2+3, etc... step for the other group of 5 stacks....

Hypo · 03-15-2002, 02:29 AM

Hey, so I was right! Thank you, thank you. *Takes a Bow*

03-14-2002, 08:14 AM	#1
Pipecrew Master of Gfy.com Industry Role: Join Date: Feb 2002 Posts: 14,887	Puzzle for the Smart Kids :) There are ten seemingly identical stacks, each containing ten coins. One of these stacks, however, in its entirety, is counterfeit and weighs differently from the genuine coins, presumed to be only detectable by employing a scale. You are challenged to find the absolute minimum number of weighings necessary to indisputably locate that counterfeit stack. Be advised, that any and each weight reading is considered a separate weighing. The weights of the two types of coins may be assumed, whereby you can facilitate simplicity in mathematics. Your answer should include an explanation of your manipulation of the coins, in order to justify the minimum number of weighings in your solution and encourage discussions among participants. This puzzle contains no deceptive wording. The correct strategy of solving is purely logical, totally realistic and physically executable. ========== Some guy posted this on my sites chat board, if you got any idea's post it here and I will post it, shutting him up for good (one of those guys whose life mission, is to prove he is smarter then everyone)

03-14-2002, 08:58 AM	#5
Jakke PNG ex-TeenGodFather Industry Role: Join Date: Nov 2001 Location: Suomi Finland Perkele Posts: 20,306	Hehe... just weight each stack once The one with different weight is wrong. __________________ ..and I'm off.

03-14-2002, 09:04 AM	#7
Jakke PNG ex-TeenGodFather Industry Role: Join Date: Nov 2001 Location: Suomi Finland Perkele Posts: 20,306	My method was better. I'd do it that way. __________________ ..and I'm off.

03-14-2002, 10:31 AM	#15
ADL Colin Too lazy to set a custom title Join Date: Feb 2001 Location: Tube Titans, USA Posts: 11,929	Do we know if the counterfeit coins are lighter or heavier than normal? Or that has to be deduced? __________________ Adult Date Link - $50 PPS starting NOW! -- good and JUICY! skype = "adultdatelink" Last edited by ADL Colin; 03-14-2002 at 10:33 AM..

03-14-2002, 10:32 AM	#16
ADL Colin Too lazy to set a custom title Join Date: Feb 2001 Location: Tube Titans, USA Posts: 11,929	A balancing scale or a regular scale? __________________ Adult Date Link - $50 PPS starting NOW! -- good and JUICY! skype = "adultdatelink"

03-14-2002, 08:22 AM	#2
Za Ha Confirmed User Join Date: Oct 2001 Location: Still lost Posts: 5,112	Too much to read But answer the question in my signature.... now thats hard.

03-14-2002, 08:26 AM	#3
Living For Today Confirmed User Join Date: Feb 2002 Location: Brisbane, Australia Posts: 3,970	I started attempting to answer this but my brain just cant be bothered

03-14-2002, 08:55 AM	#4
Pipecrew Master of Gfy.com Industry Role: Join Date: Feb 2002 Posts: 14,887	haha well if anyone gets motivated or can read that much answer the damn thing

03-14-2002, 09:01 AM	#6
Hypo So Fucking Banned Join Date: Feb 2001 Posts: 1,104	4 weightings, 5 max. First weight 4 and 4. If they are equal then one of the two remaing stacks is faulty. If they are unequal separate one group of 4 into 2 n 2 and weight - if equal the other 4 is faulty. Separate those into 2 and 2. Weight. Weight. Thats about it. You can pretty much figure out all the options.

03-14-2002, 09:07 AM	#8
NetRodent Confirmed User Join Date: Jan 2002 Location: In the walls of your house. Posts: 3,985	I would need 3 to 5 attempts at weighing the stacks of coins to find the counterfeit stack. 1. Pick 1 stack of coins and exclude it. Weigh the remaning 9 stacks of coins. 2. Divide the nine stacks into 3 sets of 3 stacks and weigh two of them. The third stack I deduce the weight of by subtracting the weight of the two known stacks from the measurement in step 1. If I am lucky all 3 sets of 3 stacks weigh the same meaning the stack I excluded in step one is the counterfeit stack. 3. If not, at least I know which of the sets the counterfeit stack is in. From that set I would pick any stack at random and weigh it. By this point by comparing the weights I measured in step 2 I know what a stack of real coins should weigh. If the weight is not proper for a real stack of coins, I've found it on the 4th weighing. If it is proper, I take one of the two remaining stacks and weigh it. This fifth weighing is the last one I need. If the stack wieghs the correct amount, I know the other stack is counterfeit. If It doesn't weigh the correct amount then it is the counterfeit stack. Thus 3-5 weighings are all that is needed.

03-14-2002, 09:09 AM	#9
Hypo So Fucking Banned Join Date: Feb 2001 Posts: 1,104	On second thought, I have a better answer - just 1 weighing will do! Take 1 coin from stack 1. Two coins from stack 2. Three coins from stack 3 etc. Weigh all the coins, divide the excess weight with the weight of the counterfeit coin. You get the answer.

03-14-2002, 09:13 AM	#11
NetRodent Confirmed User Join Date: Jan 2002 Location: In the walls of your house. Posts: 3,985	What sort of device are we able to use for measuring weight? I see that Hypo assumed it was a balance thus being able to compare two sets in one weighing. I assumed it was a calibrated scale allowing only one stack or set of stacks to be weighed at a time.

03-14-2002, 09:41 AM	#13
HeadPimp Bad Mo-Fo Industry Role: Join Date: Jul 2001 Location: Seattle, WA Posts: 2,772	Fuck that! Lets go with 0 weighings. The say that the coins are of different weights but of identical dimensions indicated that they are of different densities. By using a weight with a point on it, dropped from a fixed height, you can create a dimple in each coin. Teh coin that has a different size dent is the fake. OR! Arrange the stacks around the rim of a large disk that is perfectly balanced at its center. The stack that either comes out higher or lower is the fake stack.

03-14-2002, 09:45 AM	#14
FADE19 Snow's Parole Officer Join Date: Sep 2001 Location: mud hut next to Bin Laden's Posts: 1,161	1+1=3

03-14-2002, 10:38 AM	#17
ADL Colin Too lazy to set a custom title Join Date: Feb 2001 Location: Tube Titans, USA Posts: 11,929	Here .. since you wanna prove him wrong ... Fastest way to find a solution is to look it up ;) http://scientium.com/drmatrix/puzzles/mg1puz5.htm You might wanna change the wording of the answer. __________________ Adult Date Link - $50 PPS starting NOW! -- good and JUICY! skype = "adultdatelink"

03-14-2002, 10:39 AM	#18
bauhaus Confirmed User Join Date: May 2001 Location: Costa Rica Posts: 773	I think the answer is 6 at most.... 1. weigh 5 stacks record 2. weigh 5 stacks record, then continue with the lightest/heaviest (depending on what you assume the variable to be) 3. now you have 5 stacks left.... 4 separate weights are taken next ex. 1 2 3 4 5 weight of 1+2, 2+3, 3+4, 4+5 4. now you have taken 6 weights.....out of the 4 weights above you can figure out which one is heaviest/lightest (again depending on what you set you known variables to be) using comparitive measures.... Unless it is a lateral thinking trick....it can't be less ie...you can use a counterbalanced scale (therefore you could balance the scale against the different stacks, thereby reducing the amount of total weights needed) __________________ Insane, Color and Instant Sensation REalize the potential of dynamic coded features all in real time Vivid Php Code This is the reality of Outsoursouring SKype HolyBauhaus Deep, Deep in the Tropical Dry Rainforest.

03-14-2002, 10:46 AM	#19
ADL Colin Too lazy to set a custom title Join Date: Feb 2001 Location: Tube Titans, USA Posts: 11,929	The answer is one. Very clever problem. __________________ Adult Date Link - $50 PPS starting NOW! -- good and JUICY! skype = "adultdatelink"

03-14-2002, 10:53 AM	#20
bauhaus Confirmed User Join Date: May 2001 Location: Costa Rica Posts: 773	The weights of the two types of coins may be assumed, whereby you can facilitate simplicity in mathematics. If this is assumed then you could pick the right one first? that would give you an answer of one? Colin how else? __________________ Insane, Color and Instant Sensation REalize the potential of dynamic coded features all in real time Vivid Php Code This is the reality of Outsoursouring SKype HolyBauhaus Deep, Deep in the Tropical Dry Rainforest.

03-14-2002, 10:55 AM	#21
FADE19 Snow's Parole Officer Join Date: Sep 2001 Location: mud hut next to Bin Laden's Posts: 1,161	CRAP....I was soooo close I had 3....damn it took me awhile to come up with that formula too.

03-14-2002, 10:57 AM	#22
ldinternet Confirmed User Join Date: Apr 2001 Posts: 8,245	Colin, the given answer has its limitations in that it assumes that you can choose the mass overage for the counterfeit coins. How about having a standardised overage of mass for the counterfeit coins of, for example, 0.02kg?

03-14-2002, 11:02 AM	#24
bauhaus Confirmed User Join Date: May 2001 Location: Costa Rica Posts: 773	ID I see what you are saying...but in your example of a difference of .02kg....its easy because the the stacks are in groups of 10 and the kilograms are metric...you would have mind fucked us if you used your fuckin imperial weights.. If you were not sure of anything my way would work using the minimun amount of weighs.....if you could not even assume the counterfeits were more or less you you just have to do the 1+2, 2+3, etc... step for the other group of 5 stacks.... __________________ Insane, Color and Instant Sensation REalize the potential of dynamic coded features all in real time Vivid Php Code This is the reality of Outsoursouring SKype HolyBauhaus Deep, Deep in the Tropical Dry Rainforest.

03-15-2002, 02:29 AM	#25
Hypo So Fucking Banned Join Date: Feb 2001 Posts: 1,104	Hey, so I was right! Thank you, thank you. Takes a Bow