Using the Pigeon-hole principle...

91 five-digit numbers are written on a blackboard. Prove that one can find three numbers on the blackboard such that the sums of their digits are equal.

I don't know what the pidgeon-hole principle is, but I imagine the solution will go something like this:

The largest five-digit number is 99999, the sum of whose digits is 45.
The smallest five-digit number is 10000, the sum of whose digits is 1.

You can find any sum between those values by some choice of 45 five-digit numbers, so if you choose 91 such numbers, then whatever they are three of them must have the same sum, because there are two lots of 45 numbers plus 1 in 91.

To prove that there will always be three numbers on the blackboard with equal sums of digits, we can use the Pigeonhole Principle.

Let's consider the possible sums of digits for a five-digit number. The sum of digits can range from 0 (if all the digits are 0) to 45 (if all the digits are 9). So, in total, there are 46 possible sums of digits (0 to 45) for a five-digit number.

Now, let's assume that all the 91 five-digit numbers on the blackboard have distinct sums of digits. In that case, the pigeonhole principle tells us that at most, we can have 45 numbers with different sums of digits since there are only 46 possible sums of digits.

However, we have 91 numbers on the blackboard, which is greater than 45. By the pigeonhole principle, there must be at least two numbers that have the same sum of digits.

Now, let's consider the worst-case scenario where we have exactly two numbers with the same sum of digits. We still have one more number to consider.

Since the sum of digits can range from 0 to 45, and we have already assigned two numbers with the same sum of digits, the remaining number must have a sum of digits that is equivalent to one of the other numbers. Therefore, we have found three numbers on the blackboard such that the sums of their digits are equal.

This proves that, regardless of the actual arrangements of the numbers, there will always be at least three five-digit numbers on the blackboard with equal sums of digits.