Solution: The rigged scale

We publish here the solution to the divertimento The rigged scale.

The fun:

A trader sells rice using a scale with one arm longer than the other and a single 1 kg weight.

In order not to arouse suspicion, she switches the weight from one side to the other after each sale, using the fault in the scale on the first transaction of each day to her advantage. “So,” she says to herself, “if I sell 10 kg of rice in one morning I don’t make any additional profit, but if I sell 11 kg I make something on the last transaction. It’s not much, but it’s not good to be too greedy either, and business is not so bad: every day I sell some rice.

There are days when, indeed, this ruse does not bring her any profit or loss. Prove that the ratio between the lengths of the arms of the scales is a rational number.

The solution:

Let \(a\) and \(b\) be the lengths of the arms of the scale, with \(a > b\), and let \(M\) be the price of 1 kg of rice. We begin by observing that the trader, contrary to what she thinks, loses money with every two consecutive sales in which she shifts the weight. According to the law of the lever, in each sale with the scale in her favour she sells rice worth \(Mb/a\) € and receives \(M\)€, and in each sale with the scale against her she sells rice worth \(Ma/b\) € and receives \(M\)€. Therefore, the difference between the value of the rice he delivers and the money he receives is always positive:
$$\frac{b}{a}M + \frac{a}{b}M – 2M = M \frac{(a-b)^2}{ab} >0$$
It follows that the trader is harmed by this practice on days when she sells rice an even number of times. Therefore, on a day when she is neither disadvantaged nor advantaged she necessarily makes an odd number of sales \(2N+1\), with \(N >1\). If the disadvantage caused by the balance in the first \(2N\) sales equals the advantage of the last sale, the following equality must be satisfied
$$N M \frac{(a-b)^2}{ab} = M – M\frac{b}{a},$$
therefore
$$N \frac{(a-b)^2}{ab} = \frac{a-b}{a}, \qquad N= \frac{ab(a-b)}{(a-b)^2a} =\frac{b}{a-b}. $$
Then,
$$\frac{1}{N} = \frac{a-b}{b} = \frac{a}{b}-1 , \qquad  \frac{a}{b} = 1+ \frac{1}{N} \in \mathbb{Q}.
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