6. Let a composition contain $n>1$ wagons with gold coins. There are two types of visually identical gold coins: genuine and fake. In each wagon, there are only coins of one type. Coins of the same type have the same mass, while coins of different types do not have the same mass. The mass of a genuine gold coin is known.

Determine the minimum number of measurements on a digital scale that can determine which wagons contain fake gold coins, as well as the mass of a fake gold coin.

(It is assumed that any number of coins can be taken from each wagon.)

(Miloš Milosavljević)
6. We will prove that the minimum number of measurements is equal to 2. Let the weights of the genuine and counterfeit gold coins be denoted by $x$ and $y$ respectively, and let $a_{i}=1$ if the gold coins in the $i$-th wagon are counterfeit, and $a_{i}=0$ otherwise.

In the first measurement, take one gold coin from each wagon. Then we have $a_{1} + a_{2} + \cdots + a_{n} = \frac{n x - m_{1}}{x - y}$, where $m_{1}$ is the obtained weight. Assume that $m_{1} \neq n x$, otherwise there are no counterfeit coins. In the second measurement, for some $q \in \mathbb{N}$, take $q^{i-1}$ gold coins from the $i$-th wagon. If the obtained weight is $m_{2}$, we have $a_{1} + q a_{2} + \cdots + q^{n-1} a_{n-1} = \frac{\left(1 + q + \cdots + q^{n-1}\right) x - m_{2}}{x - y}$. From this, we get

$$
f\left(a_{1}, a_{2}, \ldots, a_{n}\right) = \frac{a_{1} + q a_{2} + \cdots + q^{n-1} a_{n-1}}{a_{1} + a_{2} + \cdots + a_{n}} = \frac{\left(1 + q + \cdots + q^{n-1}\right) x - m_{2}}{n x - m_{1}}
$$

We want the value of $f$ to uniquely determine $a_{1}, \ldots, a_{n}$. Therefore, it is sufficient to show that there exists a natural number $q$ such that the function $f: \{0,1\}^{n} \backslash \{(0,0, \ldots, 0)\} \rightarrow \mathbb{R}$ is injective.

For fixed $\mathfrak{a} = \left(a_{1}, a_{2}, \ldots, a_{n}\right)$ and $\mathfrak{b} = \left(b_{1}, b_{2}, \ldots, b_{n}\right)$, the equality $f(\mathfrak{a}) = f(\mathfrak{b})$ is equivalent to $P_{\mathfrak{a}, \mathfrak{b}}(q) = \left(a_{n} b - b_{n} a\right) q^{n-1} + \cdots + \left(a_{2} b - b_{2} a\right) q + \left(a_{1} b - b_{1} a\right) = 0$, where $a = a_{1} + a_{2} + \ldots + a_{n} \neq 0 \neq b_{1} + b_{2} + \ldots + b_{n} = b$. Therefore, if the function $f$ is not injective, then $q$ is a root of the polynomial

$$
P(q) = \prod_{\mathfrak{a}, \mathfrak{b}} P_{\mathfrak{a}, \mathfrak{b}}(q)
$$

Since none of the polynomials $P_{\mathfrak{a}, \mathfrak{b}}$ are identically zero, there are only finitely many numbers $q$ for which $P(q) = 0$, so it is possible to choose $q$ such that $f$ is an injective function.

Thus, in two measurements, we can determine $a_{1}, \ldots, a_{n}$, i.e., the wagons with counterfeit gold coins. Finally, $y$ is determined from the equality $y = x - \frac{n x - m_{1}}{a_{1} + a_{2} + \cdots + a_{n}}$.

On the other hand, one measurement is not sufficient, because by taking $k_{i}$ gold coins from the $i$-th wagon, we get the equation $k_{1} a_{1} + \cdots + k_{n} a_{n} = \frac{k x - m}{x - y}$ (where $k = k_{1} + \cdots + k_{n}$), which in general has multiple solutions. For example, two possible solutions for $\left(a_{1}, a_{2}, \ldots, a_{n}, y\right)$ are $\left(1, 0, \ldots, 0, x - \frac{k x - m}{k_{1}}\right)$ and $\left(0, 0, \ldots, 1, x - \frac{k x - m}{k_{n}}\right)$.