First, there are two key realizations you need before you can solve the problem:

1. You may multiply the weights of all the masses by any constant and not have any effect on the puzzle

2. You may add any constant to the masses and not have any effect on the puzzle

The proofs of these are pretty simple, so I won't bother to do it.

So, first thing is that you use your power of a multiplicative constant to give all the masses integer mass (keep multiplying by denominators until you are there, basically). Next, you can use the additive constant to make one of the masses zero. Finally, you can halve all the masses as many times as it takes for there to be at least one weight with an odd mass (this step fails iff all the weights have the same mass). So, you have 11 integer masses and at least one of them is zero and at least one of them is odd.

Now, remove the zero mass, the remaining ten can be divided into two groups with the same (integer) mass. Thus, the total mass remaining must be an even number. Thus the total mass even with the zero mass included must be an even number.

Now, remove the odd mass, the remaining ten can be divided into two groups with the same (integer) mass. But, the total mass of the weights left behind must be an odd number. This is a contradiction, proving the masses must all be the same.

In the case where the masses can be real, I believe the proof goes along the same lines, but I do not actually know it myself.

## 1 comment:

That is a cute proof. Even after reading it, I had to think about it though.

!bob

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