p1. Consider the sequence of positive integers $2, 3, 5, 6, 7, 8, 10, 11 ...$. which are not perfect squares. Calculate the $2019$-th term of the sequence. p2. In a triangle $ABC$, let $D$ be the midpoint of side $BC$ and $E$ be the midpoint of segment $AD$. Lines $AC$ and $BE$ intersect at $F$. Show that $3AF = AC$. p3. Find all positive integers $n$ such that $n! + 2019$ is a square perfect. p4. In a party, there is a certain group of people, none of whom has more than $3$ friends in this. However, if two people are not friends at least they have a friend in this party. What is the largest possible number of people in the party?
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Tags: algebra, geometry, number theory, combinatorics, chilean NMO
JanHaj
27.06.2023 23:09
p1. Firstly we find the number whose square is strictly less than $2019$: $$44^2=1936 < 2019 < 2025=45^2$$So,that number is $44$. This tells us that there are $44$ perfect squares smaller than $2019.$ Now, we define the following sequences as: $$1, 2, 3, 4, 5..., 2019\quad (1)$$$$2, 3, 5, 6, 7..., 2019\quad (2)$$To get the second sequence we have simply removed the numbers which are perfect squares up to 2019 from sequence $(1)$ (which like we said, there are 44), we now know that sequence $(1)$ has 2019 elements, so to make sequence $(2)$ have 2019 elements we need to add $44$ numbers after the $2019$, so: $$2, 3, 5, 6, 7..., 2019\Rightarrow 2,3,5...,2019,2020,2021,2022,2023,2024,\boxed{2025},2026...\quad (2.1)$$Notice that after adding those number we have passed $2025$ which is a perfect square,so it must be removed from our sequence. Now sequence $(2.1)$ has $2018$ elements so we must add $1$ more number to the sequence (for it to have 2019 elements). Thus, the 2019-th term of this sequence is $2019+44+1=\boxed{2064}$
MC413551
28.06.2023 01:51
For p2 just use mass points Make A 10 so then D is also 10 and then E is 20. B is 5 and so is C. That means F is 15 So the ratio of AF to FC is 1:2 So AF to AC is 1/3