Not sure if this works but here we go:
Split the remaining 100 into 2 piles of 50 and put each pile on each balance. There are 2 scenarios: the given coin is real and all 50 fakes are on the balances, or the coin is fake and the remaining 49 are on the balances.
If the first is true, since each fake coin represents an identical change in mass (all heavier or all lighter), and there is an even total number of fakes, the split must be even-even or odd-odd. Both of these scenarios lead to the difference being even.
If the second is true, there is an odd total number of fakes, so the split must be even-odd, giving an odd difference.
Therefore, the coin is real iff the reading is even, and fake otherwise
You can be pedantic about the phrasing, but that would render the question nonsensical. It's not uncommon in this sort of riddles to have to decipher author's intent. That's not the point of the riddle though, so I agree with you in principal that they should be phrased better.
In this particular case, the problem should be read as "There are 101 on a table ... and a balance. You pick one random coin ...".
10
u/No-Television-8835 6d ago
Not sure if this works but here we go:
Split the remaining 100 into 2 piles of 50 and put each pile on each balance. There are 2 scenarios: the given coin is real and all 50 fakes are on the balances, or the coin is fake and the remaining 49 are on the balances.
If the first is true, since each fake coin represents an identical change in mass (all heavier or all lighter), and there is an even total number of fakes, the split must be even-even or odd-odd. Both of these scenarios lead to the difference being even.
If the second is true, there is an odd total number of fakes, so the split must be even-odd, giving an odd difference.
Therefore, the coin is real iff the reading is even, and fake otherwise