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Anonymous Poster

A New Coins Problem

12/19/2008 10:34 PM

There are eight silver coins and one counterfeit that looks like a silver coin, but actually weighs slightly less than the others.By using a balance scale to compare coins (or groups of coins), how can you determine the counterfeit with two weighings?
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jdretired
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#1

Re: A New Coins Problem

12/19/2008 11:08 PM

Three groups of three, should isolate the counterfeit in two weighs.

Regards JD.
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Milo
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#6
In reply to #1

Re: A New Coins Problem

12/20/2008 11:13 PM

8 = 3X3?

milo

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#22
In reply to #6

Re: A New Coins Problem

12/21/2008 6:27 PM

8 good ones plus 1 bad one = 9.

Regards JD.
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#23
In reply to #6

Re: A New Coins Problem

12/21/2008 7:16 PM

mea culpa on the arithmetic, but given three groups of three, and two weighings, the technique is not clear. You did better readingthe problem than I.

milo

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#25
In reply to #23

Re: A New Coins Problem

12/21/2008 10:08 PM

Hi, as it was a guest I thought I'd leave the thinking to him or her? But I agree Hendrik gave a good answer, it works for either 8 or 9 coins. I'll have to be more explicit.

Regards JD.
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Dedaelus
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#33
In reply to #1

Re: A New Coins Problem

03/18/2009 10:45 AM

Absolutely correct but someone wanted an explanation but looking below I saw that Hendrik gave the explanation. This one is easy.

Try this one:

Twelve identical looking balls. One is heavier or lighter. You have the same weighing balance.

In three weighings (or tries) find the odd ball and tell whether it lighter or heavier

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talos
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#2

Re: A New Coins Problem

12/19/2008 11:49 PM

Actually lots of ways to do it.

-Groups of 2.One or two groups should weigh different.(similar to the above suggestion)

-Use a jewelers scale.Has a very low least count and can be used to measure coin weights

-Most coins can simply be identified under close observation.This would be the most simplest way.Minted coins have marks which are very unique.Numismatists have an eye for these .Ask one and hopefully he will be glad to help you out

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#3

Re: A New Coins Problem

12/20/2008 12:18 AM

Take any group of 3 in each pan

if the same put remaining 2 in the pans.

if deferent take any 2 of the lighter set in the pans

if the same the remaining 1 is the one otherwise the lightest one.

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#7
In reply to #3

Re: A New Coins Problem

12/20/2008 11:14 PM

hendrik nailed it.

milo

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#11
In reply to #3

Re: A New Coins Problem

12/21/2008 12:28 AM

GA.

But little difficult to understand. Shall elaborate it moe?

Make 3 groups

Group A: 3 coins

Group B: 3 coins

Group C: 2 coins

Now put group A in one pan and group B in another pan.

1. Either the will balance: Then group C has the required Coin.

Put one coin from group C in one pan and other coin from group in other pan. The coin which weighs less is the coin we are finding.

2. If group A and group B do not balance, one which weighs less as the coin we are searching. (Let us assume that group B is weighing less.)

Put one coin from this group (B) in one pan and other coin from the same group (B) in other pan.

2a. If these two coins balance, the coin which is not being weighed is the coin we are searching for.

2b. If If these two coins do not balance, lower weight coin is the one we are searching.
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#14
In reply to #3

Re: A New Coins Problem

12/21/2008 5:40 AM

Sorry Hendrik its 9 coins not 8?

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#27
In reply to #14

Re: A New Coins Problem

12/21/2008 11:37 PM

OK, even with 9 coins, the elaborate answer I gave remains valid.

In first stap, we will have group C of 3 coins (instead of 2 coins in my earlier post)

In first weighing, we will find out the group which weighs less.

In second stage, we sort out the coin which weighs less.
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not so smart
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#4

Re: A New Coins Problem

12/20/2008 2:47 PM

my 15 year old daughter threw that one at me last summer. the three groups of three works.it was on some kind of touch screen video game she has. I think it is a game boy advanced or something.

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tonybabnik
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#5
In reply to #4

Re: A New Coins Problem

12/20/2008 10:30 PM

I don't get it!!

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#8

Re: A New Coins Problem

12/20/2008 11:30 PM

Another good A, Hendrick.

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#9

Re: A New Coins Problem

12/20/2008 11:59 PM

I guess I don't play fair, use an Ohm meter. if it resisted it is counterfeit.

Brad

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#16
In reply to #9

Re: A New Coins Problem

12/21/2008 11:04 AM

You weigh the ohm meter?

milo

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#18
In reply to #16

Re: A New Coins Problem

12/21/2008 2:00 PM

No just the leads

Brad

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narendra54
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#10

Re: A New Coins Problem

12/21/2008 12:03 AM

you take coin 123 and balance against coin 456. if either is lower. the defective ball is in the lot. abc coins in the defective lot weigh a with b. if weight is equal c is defective. either wise which has lower wt is defective in a b.

if 123 wt=wt 456

than one of 7 8 is defective. weigh them diff whichcoin is defective.
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#12

Re: A New Coins Problem

12/21/2008 4:41 AM

"There are eight silver coins and one counterfeit that looks like a silver coin"

I don't wish to be perdantic, but it reads that there are 9 coins, 8 + 1. "and" as in "George and Mary went to the movies".

Tony

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#13
In reply to #12

Re: A New Coins Problem

12/21/2008 5:34 AM

The puzzler, as stated, is ambiguous.

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#17
In reply to #12

Re: A New Coins Problem

12/21/2008 1:00 PM

I didn't realize George was seeing Mary. What movie was it?

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#28
In reply to #12

Re: A New Coins Problem

12/22/2008 2:55 PM

Please don't muddy up the problem with facts!
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#15

Re: A New Coins Problem

12/21/2008 8:40 AM

9 coins

You make a 3 arm scale (arms 120 deg apart) and put 3 coins in each pan.

The lightest group you then put in the three pans. Solved?

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otha
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#19

Re: A New Coins Problem

12/21/2008 2:14 PM

If you weigh two sets of three coins,

assume both are equal

take two of the remaining three and weigh them.

If they are equal you are holding the light coin.

You can figure what happens if the two are unequal.

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#20

Re: A New Coins Problem

12/21/2008 2:17 PM

This is what jdretired said in so many words

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#24
In reply to #20

Re: A New Coins Problem

12/21/2008 7:28 PM

"This is what jdretired said in so many words"

But not enough words to explain it, so no GA.

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#26
In reply to #24

Re: A New Coins Problem

12/21/2008 10:11 PM

Your right.

Regards JD.
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#21

Re: A New Coins Problem

12/21/2008 2:47 PM

Simple.

1. Divide the 9 coins into groups of 3.

2. Put one group of 3 on each pan, leaving the third group off.

3. If the two groups of three are balanced, the odd coin is in the untested group. If one of the weighed groups is light, it contains the odd coin.

4. Take the group of 3 that contains the odd coin, and divide it into three separate groups, and repeat the process of comparing 2 of the coins on the balance while the third coin watches.

5. Again, the balance will tell you which coin is light, or if balanced, the third coin is light.

Ed, and QED.
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#29

Re: A New Coins Problem

12/22/2008 4:51 PM

This will only work if all of the coins were just minted because coins loose weight over time.

A few months ago I did a rough test to verify whether a precision scale was working properly using some coins that I had in my pocket. I found that coins minted in that year were accurate to about .02 grams. All older coins weighed less than the new coins of the same denomination.

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#30
In reply to #29

Re: A New Coins Problem

12/22/2008 9:20 PM

I learned quite a bit from this thread.

Read. ohm meters. three armed balances, wear of circulating coins. need for unambiguous problem statements. Need for unambiguous answers. Nicely done, All.

milo

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#31

Re: A New Coins Problem

12/24/2008 10:32 AM

one of the coins is less in weght so we must devidethe coinse o three groups a,b and c

group a and b are in three coins in each and c group had two coinse.

we use the balance forthe first time to balance group a and b if they are equal so we use the balance for the second tme to balance the wo coins in group c we will find the less weight coinse.

but if the balance from the first time didnt equal so we must take the less weight group and balance two ofit cons ifthey are eqaul o the the third ons in the group is less weight or the less weight will be fnd directly from the second balance.

this is easy one but it will bemoredifficult to solve if the question become like this:

eight coinse simillar in shape and clore but one of them is differant in weight but we dont know if it is less or higher in weight han the other how we can find it by using the balance two times only?
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#32

Re: A New Coins Problem

12/26/2008 10:48 AM

Boxes and Coins: Newsletter Challenge (06/12/07)

Kenny T had a 1 weighing solution that didn't involve a scale.

Answer 212 shows a picture.

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