Lots of people have heard the puzzle where you must identify a false coin using a balance, this is a slight twist on the puzzle:
You have a digital balance, i.e. it will give a digital reading of which side is heavier and by how much, accurate to 0.1g.
You have 10 bags of coins, each bag contains 10 coins. All of the normal coins are 1g each. In the false bags, the coins are 1.1g each.
How many weighings are required to identify which bag contains the false coins?
Thank you. Thus, my solution is -
Very good, for 10 bags, the digital balance isn't actually required, a digital scale would work just as well.
However the follow on question is this:
What is the maximum number of bags of coins you could be given an still be able to identify the bag with the false coins in two weighings?
Each bag still contains 10 coins.
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 talzamirIt'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.
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.
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 talzamirThis 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.
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.
[/b]
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.