## Task 4 - 290734

Determine all pairs $\left(z_{1} ; z_{2}\right)$ of two-digit natural numbers $z_{1}$ and $z_{2}$ that satisfy the following conditions (1), (2), and (3)!

(1) It holds that $z_{1}>z_{2}$.

(2) The difference between the numbers $z_{1}$ and $z_{2}$ is 59.

(3) The difference that results when the cross sum of the number $z_{1}$ is subtracted from the cross sum of the number $z_{2}$ is 14.
I. If a pair $\left(z_{1} ; z_{2}\right)$ of two-digit natural numbers satisfies conditions (1), (2), and (3), then:

If $a, b$ are the digits of $z_{1}$ and $c, d$ are the digits of $z_{2}$, then $z_{1}=10 a+b, z_{2}=10 c+d$, $1 \leq a \leq 9,1 \leq c \leq 9,0 \leq b \leq 9,0 \leq d \leq 9$,

and from (1), (2), and (3) it follows that $10 a+b-10 c-d=59$, (4) $a+b-c-d=14$ (5)

Subtracting (5) from (4), we get $a-c=5$; (6) from this and (5) it follows that $b-d=9$. (7)

Since $b \leq 9, d \geq 0$, (7) is only possible with $b=9, d=0$ (8). Since $c \geq 1$ and (6), $a \geq 6$; from this and (6), the only possibilities for $a$ and $c$ are

$$
a=6, c=1 ; a=7, c=2 ; a=8, c=3 ; a=9, c=4
$$

With (8) and (9), it is shown that only the pairs $\left(z_{1} ; z_{2}\right)=(69 ; 10),(79 ; 20),(89 ; 30),(99 ; 40)$ can satisfy conditions (1), (2), (3).

II. These pairs satisfy these conditions, as is evident from $69>10,79>20,89>30,99>40,69-10=79-20=$ $89-30=99-40=59,(6+9)-(1+0)=(7+9)-(2+0)=(8+9)-(3+0)=(9+9)-(4+0)=14$.