[rookie1ja]

Weighing I. - Back to the Water and Weighing Puzzles
You have 10 bags with 1000 coins each. In one of the bags, all coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 gram.
If you have an accurate scale, which you can use only once, how can you identify the bag with the forgeries? And what if you didn't know how many bags contained counterfeit coins?

Spoiler for Solution:

Weighing I. - solution
If there is only 1 bag with forgeries, then take 1 coin from the first bag, 2 coins from the second bag ... ten coins from the tenth bag and weigh the picked coins. Find out how many grams does it weigh and compare it to the ideal state of having all original coins. The amount of grams (the difference) is the place of the bag with fake coins.
If there is more than 1 bag with forgeries, then there is lots of possible solution. I can offer you this one as an example: 1, 2, 4, 10, 20, 50, 100, 200, 500 and 1000.

Spoiler for old wording:

Imagine you have 10 bags full of coins, in each bag 1000 coins. In one bag, there are all coins forgeries. The original coin is 1 gram light, forgery is 1.1 gram. Balancing (Edit: Weighing) just once on an accurate weighing-machine, how can you identify the bag with forgeries? And what if you didn't know how many bags contain forgeries?

[kola]

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.

[vinay0007]

Binary series is more precise.. which depicts.
2^0, 2^1, 2^2.... so on.
and whatever the extra weight it will be simple for the calculation..
100101010 format

[Garrek99]

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.

[Baikin]

There's a mistake on that sequence, so there's also a mistake on proving it wrong.

The sequence should be:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, and so on. This sequence should meet the mentioned conditions.

[Camshaft]

Place all bags on the scale, it should read 10,100g. Take off a bag until the 100 at the end goes away. That will be the bag with the forgeries. If you don't know how many bags contain the forgeries, then still place all bags on the scale and every time 1,100g is removed from the weight that bag will be full of forgeries.

[Garrek99]

Quote

Place all bags on the scale, it should read 10,100g. Take off a bag until the 100 at the end goes away. That will be the bag with the forgeries. If you don't know how many bags contain the forgeries, then still place all bags on the scale and every time 1,100g is removed from the weight that bag will be full of forgeries.

You are given one opportunity to use the balance. Your method requires 1 or more tries.

[puppyluv1725]

you know
all these weird riddles and stuff make my brain hurt like heck
can you make them a bit clearer and not so hard

[pilgrim]

It's a good one

[luioliher]

You can always open the bag