Count Distinct Numbers on Board

Too many down votes. Because this is a brain teaser.

The correct answer is that if n is 1, the only integer on the board will be 1 itself. For any n > 1, there will always be n distinct integers, including n, n - 1, down to 1. This is because each day, the integer n - 1 gets added to the board, then n - 2, and so on until 1.

This makes the correct Python solution as follows:

class Solution:
    def distinctIntegers(self, n: int) -> int:
        # If n is 1, the only integer on the board will be 1 itself.
        if n == 1:
            return 1
        # Otherwise, the integers will be n, n - 1, and 1.
            return (n-1)

The problem is a brain teaser that involves the concepts of mathematical modulos and iteration over a long duration of time (10^9 days).

Given an initial number n on a board, the task is to place certain numbers on the board each day for a total of 10^9 days. The numbers to be placed on the board each day are determined by this rule: For each number x currently on the board, you find all numbers 1 <= i <= n such that x % i == 1 and place those numbers on the board.

The insight here is that the number of days (10^9) is so large compared to the maximum possible value of n (which is 100), it’s enough time to eventually get all numbers from n down to 2 onto the board. Here is why:

Starting with n on the board, on the next day, you add n - 1 to the board because n % (n - 1) == 1 (as long as n is greater than 2). The following day, you add n - 2 because (n - 1) % (n - 2) == 1, and so forth. This continues until you add 1 to the board.

So, after an ample number of days, for any n > 1, all numbers n, n - 1, n - 2, ..., 2 will eventually be on the board, meaning there will be n unique numbers in total. Therefore, in this case, you return n.

In the case where n = 1, since there is no other i such that 1 % i == 1, only 1 itself will be on the board even after 10^9 days. Hence, in this case, you return 1.

So, the final solution is essentially to return n because for n > 1, the number of unique numbers on the board after 10^9 days will be n, and for n = 1, it will be 1.

This also shows how a very complex problem can be solved with an understanding of the underlying patterns and mathematical properties, leading to a very simple solution.