Each gear weighs either an even or an odd number of grams. Any set of 12 gears can be divided into two groups of equal weight, so a set of 12 gears has a total weight that’s even. This weight remains even if one of the 12 gears is exchanged with the 13th gear, and this is true no matter which of the 12 is exchanged. So either all the gears are of even weight or all are of odd.

Now find the weight of the lightest gear and subtract it from all 13 gears. This gives us a set of 13 gears that still fulfill the conditions of the problem (though one or more of them now has zero weight). But now we know that each gear has an even weight: If all had originally been even, then we’ve subtracted an even number to produce an even number. And if all were originally odd, then we’ve subtracted odd from odd to produce even.

Now, since each gear has even weight, we can divide the weight of each gear by 2 to obtain another set of 13 gears that still fulfills the conditions of the problem. If all the original gears had the same weight, we can continue to divide by 2 repeatedly without a problem: With each division, we produce a new set that satisfies the conditions, and every gear’s weight will have the same parity, even or odd. But if the original group of gears contained gears of differing weights, then repeated division by 2 will eventually produce a situation in which some weights are even and some odd. No such set can satisfy the original conditions of the problem. So all the original gears must have had the same weight.