On the board, only the numbers $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$, and $\frac{1}{6}$ are written. We can write on the board the sum or the product of any two numbers from the board. Is it possible to achieve by such writing that the following numbers appear on the board: a) $\frac{1}{60}$, b) $\frac{2011}{375}$, c) $\frac{1}{7}$?

(V. Bachratá, J. Mazák)
a) Yes, we can write the number $\frac{1}{60}$ on the board. For example, we can add the number $\frac{1}{10}$, which we obtain as the product of numbers already written: $\frac{1}{2} \cdot \frac{1}{5}$. Then we can write the number $\frac{1}{60}$, since it is equal to the product $\frac{1}{6} \cdot \frac{1}{10}$.

b) Yes, we can write the number $\frac{2011}{375}$ on the board. First, we will show that we can write the number $\frac{1}{375}$ on the board. This can be factored into the product of numbers that are already on the board from the beginning: $\frac{1}{375} = \frac{1}{3} \cdot \frac{1}{5} \cdot \frac{1}{5} \cdot \frac{1}{5}$. We see, therefore, that we can sequentially add the numbers $\frac{1}{15} = \frac{1}{3} \cdot \frac{1}{5}$, $\frac{1}{75} = \frac{1}{5} \cdot \frac{1}{15}$, and $\frac{1}{375} = \frac{1}{5} \cdot \frac{1}{75}$. To the number $\frac{1}{375}$, we can add another such number, specifically as the product of already written numbers 1 and $\frac{1}{375}$. By adding the numbers $\frac{1}{375}$ and $\frac{1}{375}$, we get $\frac{2}{375}$, and by repeatedly adding $\frac{1}{375}$, we can reach any fraction that has 375 in the denominator and a natural number in the numerator. Thus, we can reach the desired number $\frac{2011}{375}$.

c) No, we cannot write the number $\frac{1}{7}$ on the board. Only addition and multiplication of fractions are allowed. Let's see how these operations affect the denominator. (We will ignore that during these operations, we can also simplify fractions. This neglect will not affect our conclusion.)

- When multiplying two fractions, the denominator of the result is the product of the denominators of the original fractions.
- When adding two fractions, the denominator of the result is the product, or the least common multiple, of the denominators of the original fractions.

In the prime factorization of the product, or the least common multiple, of two numbers, there cannot be a prime number that was not in the prime factorization of either of the original two numbers. Therefore, no matter what allowed operations we perform, we will never get a denominator whose prime factorization includes a prime number that was not in the factorization of any previously written denominator. Since none of the denominators we have available at the beginning have 7 in their prime factorization, we cannot reach $\frac{1}{7}$.

Evaluation. In part a), award 1 point for a justified answer, which can also be awarded for the statement: "Yes, because $\frac{1}{60} = \frac{1}{2} \cdot \frac{1}{5} \cdot \frac{1}{6}$."

In part b), award 2 points for describing any procedure leading to $\frac{2011}{375}$. (If such a procedure is not described, 1 point can be awarded for the observation that any natural number can be obtained in the numerator due to addition.)

In part c), award a total of 3 points, of which 2 points for explaining the effect of multiplication and addition on the denominator, and 1 point for stating that multiplication and least common multiples do not yield a new prime number.