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Pipecrew

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#1

Puzzle for the Smart Kids :)

03-14-2002, 06:14 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)
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Za Ha

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#2

03-14-2002, 06:22 AM

Too much to read

But answer the question in my signature.... now thats hard.
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Living For Today

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#3

03-14-2002, 06:26 AM

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

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#4

03-14-2002, 06:55 AM

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

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#5

03-14-2002, 06:58 AM

Hehe...
just weight each stack once
The one with different weight is wrong.

..and I'm off.
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Hypo

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#6

03-14-2002, 07:01 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.
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Jakke PNG

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#7

03-14-2002, 07:04 AM

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

..and I'm off.
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NetRodent

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#8

03-14-2002, 07:07 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.

"Every normal man must be tempted, at times, to spit on his hands, hoist the black flag, and begin slitting throats."
--H.L. Mencken
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Hypo

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#9

03-14-2002, 07:09 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.
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EasyDialers

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#10

03-14-2002, 07:09 AM

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

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NetRodent

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#11

03-14-2002, 07:13 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.

"Every normal man must be tempted, at times, to spit on his hands, hoist the black flag, and begin slitting throats."
--H.L. Mencken
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NetRodent

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#12

03-14-2002, 07:14 AM

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?

Last edited by NetRodent; 03-14-2002, 07:44 AM.

"Every normal man must be tempted, at times, to spit on his hands, hoist the black flag, and begin slitting throats."
--H.L. Mencken
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HeadPimp

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#13

03-14-2002, 07:41 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.
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FADE19

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#14

03-14-2002, 07:45 AM

1+1=3
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ADL Colin

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#15

03-14-2002, 08:31 AM

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

Last edited by ADL Colin; 03-14-2002, 08:33 AM.

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ADL Colin

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#16

03-14-2002, 08:32 AM

A balancing scale or a regular scale?

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ADL Colin

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#17

03-14-2002, 08:38 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.

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bauhaus

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#18

03-14-2002, 08:39 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)

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ADL Colin

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#19

03-14-2002, 08:46 AM

The answer is one.

Very clever problem.

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bauhaus

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#20

03-14-2002, 08:53 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?

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FADE19

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#21

03-14-2002, 08:55 AM

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

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#22

03-14-2002, 08:57 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?
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#23

03-14-2002, 08:57 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.....

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#24

03-14-2002, 09:02 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....

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Hypo

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#25

03-15-2002, 12:29 AM

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