19 replies to this topic

[rookie1ja]

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

Weighing III. - solution
Similar to the former brain teaser.
I take out 0 (no coin from the first bag), 1 (one coin from the second bag etc.), 2, 4, 7, 13, 24, 44 coins (from the last 8th bag). Each triple is unique enabling an easy way to identify the bags with fake coins (using only 95 coins).

Spoiler for old wording

This puzzle is a step further than the previous one (weight of fake and non fake coins is the same as the weight of bears in the previous puzzle). You have eight bags, each of them containing 48 coins. Five of these bags contain only true coins, the rest of them contain fake coins. Fake coins weigh 1 gram less than the real coins. You do not know what bags have fake coins and what bags have real coins. You can use a scale, a dynamometer type one, with precision up to 1 gram (an accurate weighing machine). Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins?

[Jawsh]

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.

[Martini]

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.

[soldier3001]

I think I found the solution:

Spoiler for ...

Assign numbers to each of the bags such that the sum of any 3 numbers assigned to the bags does not equal the sum of any other three bags. Take the number of coins that corresponds to the number on the bag. Put them all in an individual pile on the scale and push the button and record the number on the scale. With simple math you can find which bags' coins are causing the difference in weight compared to what it would be if all the coins were real.
For instance:
Bag 1: 0
Bag 2: 1
Bag 3: 2*
Bag 4: 3
Bag 5: 7*
Bag 6: 13
Bag 7: 24
Bag 8: 45*
In this scenario, if bags 3, 5, and 8 are fake, there would be a difference of 54 grams registered on the scale. We would know that only 45, 7, and 2 add together to equal 54, so it must be bags 3, 5, and 8. I don't know if this is the minimum, but I think it is.

[spoxjox]

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.

[qvivid]

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

[qvivid]

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

[spoxjox]

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).

[rahul]

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?

[rahul]

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.