Refraining from giving the answer.. would love to know how to make the hidden text thingy. The solution I have in mind works for 11 bags of 10 coins too.

Only one weighing, with a pile consisting of 1 coin from the 1st bag, 2 from the second etc, all the way to 10 from from the 10th. There are 55 coins, of which 1 .. 10 are false, making a total of 55 + n/10 grams of combined weight where n is the ordinal of the bag where the heavier false coins are. If there are 11 bags, no coins from the last bag mean that 55g total weight implies that the bags in the 11th coin are false.

I formed a square of 11 rows and 11 lines of the bags, numbered 0...10. First weighing has 0 coins from each bag on line 0, 1 from each on line 1, etc, 10 from each on line 10. Second weighing does the same for row 0, row 1, etc. On each weighing, there are 11 x (0 + 1+ 2+ ... + 10) coins weighed, which is 565g of coins; but false coins cause the weight to be a bit over that. Subtract 565g from each value and multiply by 10 gives integer values m and n, and those are the coordinates of the bag with the false coins.

I wonder if we refer to a different kind of measuring device. Digital balance scales that I know of simply have stuff put on top and give an accurate measurement of the weigh on a display, usually a LED.

Originally posted by talzamir I wonder if we refer to a different kind of measuring device. Digital balance scales that I know of simply have stuff put on top and give an accurate measurement of the weigh on a display, usually a LED.

It's possible my terminology is wrong or nonsensical. I meant a two-pan balance, with a digital readout of which side is heavier and by how much.

Similar to this: http://www.scale-digital.net/wp-content/uploads/2011/08/Balance-Scale.jpg
except the slider part is digital.

This is very possibly a theoretical device; sorry for the confusion.

Originally posted by talzamir I formed a square of 11 rows and 11 lines of the bags, numbered 0...10. First weighing has 0 coins from each bag on line 0, 1 from each on line 1, etc, 10 from each on line 10. Second weighing does the same for row 0, row 1, etc. On each weighing, there are 11 x (0 + 1+ 2+ ... + 10) coins weighed, which is 565g of coins; but false coins cause the weight ...[text shortened]... 0 gives integer values m and n, and those are the coordinates of the bag with the false coins.
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This solution seems sound. It took me a second to follow the bolded statement, since you can't include 0 coins from multiple bags in a weighing.

You can, however, set aside 11 bags and weigh 11x (1 + 2 + ... + 10) coins from 110 other bags, which is I think what you meant and provides the same conclusion of 121 bags.