Milikan's Experiment Lab - Euclidean Algorithm

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The discussion revolves around a lab experiment similar to Millikan's, where the goal is to determine the mass of a single jellybean from various bags with different total masses. The user proposes using the Euclidean algorithm to find the greatest common divisor (GCD) of the adjusted masses after subtracting the bag weights. A challenge arises in knowing when to stop the algorithm since it can theoretically continue indefinitely with real numbers. The suggested approach involves repeatedly applying integer division and tracking remainders until only one non-zero remainder remains, which indicates the mass of one jellybean. This method effectively narrows down the possible mass of the jellybean through systematic calculations.
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Homework Statement



Hey guys I need help with a lab I'm doing that is similar to the Milikan's experiment. I am given 10 bags each holding the same item (Jellybean) of various quantities. Each bag has a different mass. What I'm trying to figure out is the mass of the individual item, so mass of 1 jellybean. This is similar to finding the elementary charges for milikan's experiment.

Homework Equations


I believe using the euclidean algorithm is the best method because all the bags should be a integer multiple of the mass of 1 jellybean.


The Attempt at a Solution


First i subtracted the mass of the bags from the mass of bag + jellybeans, to find the mass of the jellybeans only. Then i ordered the various masses in increasing order. Now my problem is, in doing Euclidean Algorithm to find GCD, since these numbers are real numbers, how do I know when to stop using the algorithm. I've made a Java program that does the algorithm recursively but I'm not sure when to stop because this algorithm can go on forever. Thanks in advance.
 
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Do integer division with the smallest mass (m1) (result is Ni for the i-th mass) and replace all masses by the remainder mi-Ni*m1.Choose the smallest again and repeat till you get zero remainder for all but one. The last smallest mass still can include a few beans, but that is the best you can do.

As an example: assume that here are 5 bags of mass 33, 16.83, 15.84, 5.94, 2.31. The smallest is 2.31, integer division gives 14, 7, 6, 2, and the remainders are 0.66, 0.66, 1.98, 1.32, Choose these values for mi and add also 2.31. The smallest is 0.66. Repeat the procedure, the remainders are 0, 0, 0, 0, 0.33. You have one non-zero remainder, that is the mass of one bean (or integer multiple of it).

ehild
 
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