Consider the four-digit numbers in which neither 4 nor 5 appear as digits. Form the sum of any two such different numbers. How many different numbers appear among the sums?
A two-word sum is equal to each other if and only if the sum of the digits in each column is the same in the usual addition scheme. This is obviously true for the last (the $10^{0}$ place value) column, since the sum in it can only be $4+4=8, 4+5=9, 5+5=10$, meaning we cannot get the same digit to write down with or without carrying over. And if we now do not carry over the possible remainder, but write a new (tens) digit, say $X$, in each of the two equal sums (or interpret the operation in a number system with a base greater than ten, which does not change whether the sums are equal or different), then our statement also holds for the $10^{1}$ place value column, and so on for all 4 columns.

Accordingly, we can think of as many sums as there are four-digit numbers that can be written with the digits $8, 9, X$, that is, forming permutations with repetition from 3 different elements, 4 at a time (4th class) - $3^{4}=81$ variations.

However, since the sums can only be formed from two different terms, we must delete those sums in which every digit is 8 or $X$, because these and only these can result exclusively from equal lower and upper digits, the number of these is $2^{4}=16$. Thus, the number of different sums that meet the requirements is $3^{4}-2^{4}=65$.

Kabay György (Miskolc, Földes F. Gimn., III. o. t. )