The Market Scales Puzzle

 A two-pan scale allows the market vendor to place weights on the same pan as the herbs (subtracting its value from the amount on the opposite pan) or on the opposite pan (adding its value). This means each weight has three options for any given measurement. If the weight is placed in the opposite pan from the herbs, it adds to the weight measurement of the herbs. Not using a given weight does not affect the measurement. And placing a given weight in the same pan as the herbs subtracts from the measured weight of the opposite pan.

If we begin with the weight 1g, we can measure only 1g. But if the next weight is 3g, we can measure from 2g (using 3g - 1g) to 4g (using 3g + 1g). The next weight we would need is 9g, since then we can measure from 5g (9g -3g - 1g) to 13g (9g + 3g + 1g). Finally, if we have a fourth weight of 27g, we can measure from 14g (27g - 9g - 3g -1g) to 40g (all four weights together). Thus, having weights 1, 3, 9, and 27 (powers of 3) allow us to measure any amount from 1 to 40 grams. This seems to be connected to the fact that each weight can be used in 3 ways.

If we are using a one-pan scale in which weights can only go on one side, then the weight is either used (adding its value) or not used. This means there are only two choices per weight. My hunch is that this means we would need weights that are powers of two, but let's test it. With 1g we measure 1g. So we need 2g, which allows us to measure up to 3g. Then we need 4g, which allows us to measure up to 7g. Then we need 8g, which allows us to measure up to 15g. And finally, with a 16g weight, we can measure up to 31g. So we would need to use the weights 1g, 2g, 4g, 8g, and 16g to measure up to 31g.

So the concept seems to be the following: in a one-pan scale, we are showing that every integer has a unique representation as a sum of distinct powers of 2. With the two-pan scale, we are representing every integer uniquely as some combination of powers of 3, but in a way that allows us to add or subtract (or not use) those powers.

As far as a possible extension of this puzzle, we could ask students what the next weight needed in the two-pan scale would be to measure weights beyond 40g without gaps. The answer should be 81g, or 3 to the 4th power. This would allow us to measure up to 121g. One way to look at this is to see that with 5 weights and 3 choices per weight, we have 3^5 = 243 possible combinations. But since a weight of 0 grams is useless and every positive total has a perfect negative mirror image (just swap the pans), we divide the remaining 242 combinations by 2, which gives exactly 121 unique positive weight amounts (from 1 to 121g).

All I can think of in terms of number theory is that this is a way of showing a complete partition of the integers. Basically, every integer has a unique representation as a sum of distinct powers of 3 using coefficients from the set {-1, 0, 1}.