## Task 2 - 340822

From the digits $0,1,2,3,4,5,6,7,8,9$, two fractions $\frac{a}{b}$ and $\frac{c}{d}$ are to be formed such that each digit is used exactly once in the digit representations of the four natural numbers $a, b, c, d$. For the fractions formed, $\frac{a}{b}+\frac{c}{d}=1$ should hold. In the first of the two fractions, $a=13$ and $b=26$ should be chosen.

Prove that there is exactly one way to choose the numerator $c$ and the denominator $d$ of the second fraction such that all the given conditions are satisfied! Give this second fraction!
I. If the conditions of the problem are satisfied by two natural numbers $c$ and $d$, it follows that: Since $a=13$ and $b=26$, from $\frac{a}{b}+\frac{c}{d}=1$, it follows that $\frac{c}{d}=\frac{1}{2}$, so $2-c=d$ (1) holds.

Furthermore, it follows that in the digit representations of $c$ and $d$, only the digits $0,4,5,7,8,9$ appear, each exactly once.

If $c$ were at most two digits, and thus $d$ at least four digits, then $c \leq 99, d \geq 1000$; if $c$ were at least four digits, and $d$ at most two digits, then $c \geq 1000, d \leq 99$; both in contradiction to (1).

Therefore, $c$ and $d$ are both three digits.

The hundreds digit of $c$ is 4, otherwise $d$ would be four digits due to (1). The units digit of $d$ is even due to (1), so it is either 0 or 8. If it were 8, then $c$ would have to have the units digit 9 due to (1) (and since the 4 is already the hundreds digit of $c$), but then the hundreds digit of $d$ would be neither 8 nor 9, which is in contradiction to (1) due to the hundreds digit 4 of $c$.

Therefore, $d$ has the units digit 0, and consequently $c$ has the units digit 5. The tens digit of $c$ is neither 7 nor 9; because of (1), so $2 \cdot 475=950$ or $2 \cdot 495=980$, the digit 5 or 9 would appear twice.

According to this and due to (1), only the possibility $c=485, d=970$ remains. (2)

II. These two numbers have the required digits, and they also satisfy (1), thus $\frac{a}{b}+\frac{c}{d}=1$.

With I. and II., it is shown that exactly the fraction $\frac{4850}{970}$ formed with the numbers (2) satisfies the conditions of the problem.