Hey guys,
Just found an interesting problem in the Online Mathematics Olympiad this year. It states the following:
I wanna add an alternative solution to the official one. Let the sum inside the ceiling function be A, then A can be rewritten as follows:
Let:
Then A can be written as:
It's easy to see that:
The inequality is due to the fact that for the S(n+1), every term in it is less than that of S(n) (for example, 1 over n becomes 1 over n+1, which is clearly less than 1 over n because n + 1 > n for positive n, in which we will use positive because we will plug in n = 2019 later, and I put this in terms of n because it's much easier to see the algebraic manipulations this way.)
Solving the inequality for Sn and set it for n = 2019
Plugging this back in A:
But A > 2018, because S(n) is always more than 1 (look at the expansion; it's 1 + [something], in which that [something] is obviously more than 0 so the entire S(n) > 1). Therefore, we have 2018 < A < 2019. The ceiling function rounds up any number within this range, in which the result will always be 2019. So the answer is:
😂👌2019👌😂
Just wanted to share my own way of working around this.
~Vincent
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