## Task 4 - 260734

Determine all pairs $(m ; n)$ of natural numbers that satisfy the following conditions!

(1) $m$ and $n$ are three-digit numbers.

(2) It holds that $m-n=889$.

(3) For the digit sum $Q(m)$ and $Q(n)$ of $m$ and $n$, it holds that $Q(m)-Q(n)=25$.
I. If a pair $(m ; n)$ of natural numbers satisfies the conditions, then:

Due to (1), we have $1 \leq Q(m) \leq 27, 1 \leq Q(n) \leq 27$. Therefore, (3) can only be satisfied if either $Q(m)=27, Q(n)=2$ or $Q(m)=26, Q(n)=1$.

The only three-digit number $m$ with the digit sum $Q(m)=27$ is $m=999$; for this, (2) further yields $n=m-889=110$.

The only three-digit number $n$ with the digit sum $Q(n)=1$ is $n=100$; for this, (2) further yields $m=889+n=989$.

Thus, only the pairs $(999 ; 110)$ and $(989 ; 100)$ (4) can satisfy the conditions of the problem.