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Weighing III.


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#1 rookie1ja

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Posted 30 March 2007 - 05:55 PM

Weighing III. - Back to the Water and Weighing Puzzles
This puzzle goes a step further from the previous one.
You have eight bags, each of them containing 48 coins. Five of those bags contain only true coins, the rest of them contain fake coins. A fake coin weighs 1 gram less than a real coin. You have an accurate scale, with the precision of up to 1 gram.
Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins?

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.
Pls visit New Puzzles section to see always fresh brain teasers.


Spoiler for Solution


Spoiler for old wording

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#2 Jawsh

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Posted 12 August 2007 - 10:54 PM

Since the scale only measures to 1 gram, is it logical to assume you can remove coins from the bags to measure them?

And what do you mean by you can only measure once? Measure once per bag, or once in total?
Fake coin bags should weigh 48 grams less than the real bags.
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#3 Martini

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Posted 12 August 2007 - 11:42 PM

Since the scale only measures to 1 gram, is it logical to assume you can remove coins from the bags to measure them?


Yes, it logical to assume you can remove coins from the bags.

No, the scale can not only measure up to 1 gram; it is accurate within 1 gram. Assume that it can handle weighing all coins at once.

And what do you mean by you can only measure once? Measure once per bag, or once in total?


Once in total.

Fake coin bags should weigh 48 grams less than the real bags.


Correct.
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#4 soldier3001

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Posted 21 November 2007 - 10:08 PM

I think I found the solution:
Spoiler for ...

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#5 spoxjox

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Posted 19 December 2007 - 09:49 PM

soldier3001, your reasoning is sound, but your solution is (barely) suboptimal. The original answer is the best answer, and differs from your solution only by the last number of coins being 44 instead of 45.
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#6 qvivid

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Posted 21 December 2007 - 04:05 PM

well
why not choose the solultion as below
0, 1, 2, 3, 6, 11, 20, 37
each triple is different
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#7 qvivid

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Posted 21 December 2007 - 04:25 PM

the solution I provided above is wrong
the correct answer is
0, 1. 2, 4, 7, 13, 24
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#8 spoxjox

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Posted 21 December 2007 - 07:17 PM

the solution I provided above is wrong
the correct answer is
0, 1. 2, 4, 7, 13, 24


Yes, except there are eight bags and you've only taken from seven (actually from six, taking 0 from two bags).
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#9 rahul

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Posted 22 February 2008 - 10:20 AM

the solution I provided above is wrong
the correct answer is
0, 1. 2, 4, 7, 13, 24
Yes, except there are eight bags and you've only taken from seven (actually from six, taking 0 from two bags).


I wan to understand how taking zero coins from two bags will let you identity 3 bags containing fake coins? Eg: If the total weight of the coins taken as you have mentioned, is less than the actual by 20gms. Then the solution is (13+7+(either bag1 or bag8, both of which you have omitted)). So how will you identify between bag1 and bag8?
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#10 rahul

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Posted 22 February 2008 - 10:24 AM

The solution : 1 2 3 4 7 12 21 38. The total coins taken for weighing will be only 88.

Edited by rahul, 22 February 2008 - 10:24 AM.

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