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# Weighing I.

27 replies to this topic

### #21 harpsr

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Posted 14 February 2008 - 08:54 PM

The bags of forgeries would be 100 grams lighter than the bags of genuine coins. You could determine how many bags were forgeries by wieghing them.

harpsr
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### #22 Ravi Kumar

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Posted 15 February 2008 - 05:28 AM

The solution is simple. Mark all the bags from 1 to 10. Take from the first bag only one coin, from the second take two and from the third take three and so on. At the end the result will be in the form sum of n natural numbers. Ideally the weight should be 55grams, but due to the faulty bag which ever it may be the extra portion will come into picture. For example if the first bag is the culprit then we get 55.1, if the second bag is culprit then we get 55.2 and so on.
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### #23 kolon

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Posted 13 March 2008 - 02:04 PM

You know, there is an even better solution that will make my wife happy without using a weighing machine.

1. Take all the bags of coins. Over the period of ten (or number of bags) days, give one bag to your wife and tell her to go mad, making sure that she spends every coin.
2. When the police call you to say that she has been arrested for using forged coins, you can simply take the number of days since you gave her the first bag, and you have your answer! (For more than one bag of fakes, repeat the same when she gets out of jail!)

SIMPLE!
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### #24 smiffy

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Posted 03 April 2008 - 04:59 PM

The original solution is great, but heres a tip:

Don't get the coins muddled up.
You gotta do this:

(Bag number)

1 22 333 4444 55555 666666 7777777 88888888 999999999 10101010101010101010

(Scales)

If you get them mixed up then you won't know which bag they are from:

12567562498165846582364984651893658941651983456184365182469186549

Lol
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### #25 shobith

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Posted 04 June 2008 - 05:35 AM

If we dont know how many bags has faulty coins, then min number of coins from bag i = 2 * number of coins in bag i-1.

Therefore, the sequence would be

1, 2, 4, 8, 16, 32, 64, 128, 258, 512.

Basically, represent the weight difference divided by 0.1g in binary numbers and positions of 1 will give you which bags have faulty coins.

as the coins from each bag are powers of 2, the faulty bags would give extra weights which are powers of 2 which would set the
particular binary digit.
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### #26 SilenceM

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Posted 16 September 2008 - 10:46 AM

Simply add one bag at a time to the machine and when the sum goes over 100 gr. from the 1K, that one is fake, these way you are using the machine only once & knowing exactly wish bag has fakes.
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### #27 meano

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Posted 13 October 2008 - 03:46 AM

This sequence won't work:

1, 2, 4, 8, 16, 32, 64, 128, 258, 512.

Here is why. If the 10th bag only is full of fakes you'll get an extra weigh amount of 51.2g.
And, if bags 2-9 are fake then you'll get (25.8+12.8+6.4+3.2+1.6+.8+.4+.2) 51.2g and then you won't be able to tell which situation you are in.

The total extra sum has to be distinct for any combo of fake bags.

the sequence will work

cuz the sum of both 2-9 (what u mentioned, dunno y u didnt say 1-9) and 1-9 is not 51.2. its 51 and 51.1. check it. do u know the formula for sum of a GP?
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### #28 stantonl

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Posted 17 March 2010 - 05:04 AM

To Garrek99: We are working with a scale here, not a balance. It may be used once. If you choose to weigh all ten bags, these bags must be placed on the scale, and there is no reason why they cannot be placed on the scale one at a time. As they are placed on the scale, notice how much the weight reading increases. Of course, this scale could have a blank lcd readout that will display its one and only reading to you when you hit the "ready" button. Then this would not work. Then it's back to the powers of two.
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