91 5-digit number are written on the board, how do u PROVE that there are 3 numbers on the board that the sum of the digits are equal?

The possible digit sums, for five-digit numbers, ranges from 1 (for 10000) to 45 (for 99999) That makes 45 possibilities. With 90 numbers selected, the least possible number of times any sum occurs is two, when each sum occurs twice. When you pick a 91st number, it has to be a third time for one of them.

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To prove that there are three numbers on the board with the sum of their digits being equal, we can use the Pigeonhole Principle.

The Pigeonhole Principle states that if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon. In this case, the pigeons are the 5-digit numbers and the pigeonholes are the possible sums of the digits of those numbers.

In a 5-digit number, the sum of the digits can vary from 0 (if all digits are 0) to 45 (if all digits are 9). Therefore, there are only 46 possible pigeonholes for the sums of digit values (0 to 45).

Now, let's consider the worst-case scenario to prove our claim. If each of the 91 numbers on the board has a distinct sum of digits, that means we have 91 pigeons (numbers) and only 46 pigeonholes (possible sums).

Since we have more pigeons (91) than pigeonholes (46), by the Pigeonhole Principle, at least one pigeonhole must contain more than one pigeon. In other words, at least two numbers on the board must have the same sum of digits.

To prove that there are actually three such numbers, let's assume the opposite for contradiction. Assume that there are only two numbers with the same sum of digits. In this case, there would be 90 distinct sums among the 91 numbers (one sum is repeated). However, the maximum number of distinct sums possible across 5-digit numbers is 46 (from 0 to 45).

Since the number of distinct sums (90) is greater than the maximum possible distinct sums (46), our assumption is incorrect, and there must be at least three numbers on the board with the same sum of digits.

Therefore, it is proven that among the 91 5-digit numbers on the board, there are at least three numbers that have the same sum of digits.