Faulty machine problem variation I don't know if this problem is known by any other name. The classic problem is: We have 10 machines that produce balls, each weighing 10grams. One of the machines, however, produces balls weighing 9 gramms only. We are allowed only one weighing and need to determine which machine is faulty. The solution is to number the machines from 1 to n (n=10 here) and choose k balls from k-th machine. The total weight of selected balls would be off by expected weight by k-grams. I was thinking of variations on this problem. What if we do not know by how much the faulty weight is off from the true weight, but that the difference between faulty weight and true weight is consistent? Can we still determine the faulty machine in one weighing? How about 2 weightings?

For two weightings it's quite simple:

1. Take 1 ball from every machine, `ActualWeight - 10gr*10` is the difference for the faulty machine
2. Using information from #1 you have the original task which is solvable with 1 weighting



I believe 1 weighting is not sufficient and here is why: if you take n1,n2,n3,...,n10 balls from corresponding machines, it's equally possible that 1st machine is faulty with weight difference 1/n1 and that 2nd machine is faulty with weight difference 1/n2 - these cases don't seem to be distinguishable...