a) What could be the two fractions whose sum is 8, and both are positive integer multiples of $\frac{1}{10}+\frac{1}{70}$?

b) What could be the two fractions if there is also a mixed number from which both fractions can be obtained by multiplying with an integer greater than 1?
a) Let the two fractions be $x$ and $y$, and the positive integers in the second requirement be $A$ and $B$, that is,

$$
x=A\left(\frac{1}{10}+\frac{1}{70}\right)=\frac{4}{35} \cdot A, \quad y=\frac{4}{35} \cdot B
$$

Then, according to the first requirement,

$$
\begin{aligned}
x+y=\frac{4}{35}(A+B) & =8 \\
A+B & =70
\end{aligned}
$$

We are not concerned with the order of our numbers, so it is enough to look for solutions where, for example, $x \leq y$, or $A \leq B$. Then $A$ can take the values $1, 2, \ldots, 35$, and the value of $B$ will be $70-A=69, 68, \ldots, 35$. Thus, according to (1), there are 35 pairs of fractions $x, y$ that satisfy the requirement. We have not excluded the case $A=B=35$, when $x=y=4$, because integers are also considered fractions. For example,

$$
A=1 \quad \text{in which case } \frac{4}{35} \quad \text{ and } \frac{276}{35}; \quad A=20 \quad \text{in which case} \quad \frac{16}{7} \quad \text{ and } \quad \frac{40}{7}.
$$

b) The new requirement is added to the previous two, so we can select the solutions from the above, for example, the second number example has two terms that are 2 times and 5 times the improper fraction $\frac{8}{7}$, respectively, so it meets the requirement.

Let the improper fraction in question be $r (>1)$, and

$$
x=C \cdot r, \quad y=D \cdot r
$$

where $C$ and $D$ are natural numbers greater than 1. Then, on the one hand,

$$
\begin{gathered}
x+y=(C+D) r=8 \\
C+D=\frac{8}{r} > 1
\end{gathered}
$$

In the II. and III., as well as in the V. value system, the improper fraction in question is

$$
r=\frac{8}{C+D}=\frac{8}{7}, \quad \text{or} \quad \frac{8}{5}
$$

In the VI. value system, since $A=B$, $C=D=2$, and the common value of $x$ and $y$ is 4, which can be represented by the improper fraction $2 / 1$. Therefore, the pairs of fractions that meet the new requirement are:

$$
\frac{16}{7} \quad \text{ and } \frac{40}{7}, \quad \frac{24}{7} \quad \text{ and } \quad \frac{32}{7}, \quad \frac{16}{5} \quad \text{ and } \frac{24}{5}, \quad \frac{4}{1} \quad \text{ and } \quad \frac{4}{1}
$$

Kálmán Miklós (Budapest, Fazekas M. Gymnasium II. o. t.)