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:
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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.