## Task 1 - 070921

Determine the number of all pairs of two-digit natural numbers $(m, n)$ for which $m+n=111$ holds.
The sought pairs can be divided into 2 groups:

1. Group: The sum of the units of the two two-digit numbers is 1, the sum of their tens is 11.

If the digits of a pair of numbers that meet this condition are denoted in sequence as $(a ; b),(c ; d)$, then the pairs $(a ; d),(c ; b) ;(e ; b),(a ; d)$ and $(c ; d),(a ; b)$ also meet the same condition. Because $0+1=1$ and $9+2=11 ; 8+3=11 ; 7+4=11 ; 6+5=11$, there are exactly 6 such pairs.

2. Group: The sum of the units is 11, the sum of the tens is 10.

This results in exactly 72 such pairs because $9+1=10 ; 8+2=10 ; 7+3=10 ; 6+4=10$ and $5+5=10$ for the same reason mentioned above.

In total, there are thus exactly 88 pairs of numbers that meet the conditions of the problem.

Solution adopted from $[5]$