I tried posting this twice before but could not (sorry for the inconvenience)

This relates to the balancing problem :

5 boxes , 1 box containing coins of wrong weight (correct weight = 10, wrong weight = 11)

Problem : to determine the wrong box in 1 weighing (assuming coins can be removed from boxes and that any weight can be weighed)

Generalized solution :

Let x be the correct weight and y be the wrong weight

We will consider the worst case scenario , ie where x and y are not relatively prime i.e x = 2 and y = 4

1) Choose 1 coin from the first box , 2 from the second box, 4 from the 3rd box and so on (more generally 2^(n-1) from the nth box

2) compute (1+2+4+...2^(N-1))*x , i.e (1+2+....2^(4)) * x This is the correct total weight = T , in our case

31*2 = 62

3) compute actual weight , say W

If box 1 is wrong we get y + 2x + 4x + 8x + 16x = 30x + y 30*2 + 4 = 64
If box 2 is wrong we get x + 2y + 4x + 8x + 16x = 29x + 2y = 29*2 + 8 = 66
If box 3 is wrong we get x + 2x + 4y + 8x + 16x = 27x + 4y = 27*2 + 16 = 70
If box 4 is wrong we get x + 2x + 4x + 8y + 16x = 23x + 8y = 23*2 + 32 = 78
If box 5 is wrong we get x + 2x + 4x + 8x + 16y = 15x + 16y = 15*2 + 64 = 94

4) compute number of coins from wrong box = |W-T| / |y-x| = 2^(n-1)

If box 1 is wrong we get 2^(n-1) = (64-62) / (4-2) = 2/2 , i.e n=1 If box 2 is wrong we get 2^(n-1) = (66-62) / (4-2) = 2/2 , i.e n=2 If box 3 is wrong we get 2^(n-1) = (70-62) / (4-2) = 2/2 , i.e n= 3 If box 4 is wrong we get 2^(n-1) = (78-62) / (4-2) = 2/2 , i.e n 4 If box 5 is wrong we get 2^(n-1) = (94-62) / (4-2) = 2/2 , i.e n= 5

I suppose this is an abstraction of trying to geometrically determine the center of mass / gravity of a set of points generalizable to least squares etc (if the boxes could be serially placed on a weight, the geometric configuration (sag etc) would determine the wrongly placed weight.

I suppose the above would have applications in spectroscopy / electrophoresis , error correction and other areas

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