p1. For how many natural numbers $n$, the numbers $n, 1 + 2n, 1 + 4n$ are all prime? p2. In a board of $2022 \times 2022$ squares, a piece is placed in the upper right square. Two players, Juan and Marcela, take turns, starting with Juan. On their turn, each player moves the token any positive number of squares, without leaving the board, either to the left or down. The player who places the token in the lower left corner wins. Determine which of the two players has a winning strategy. p3. Let $M$ be the midpoint of side $BC$ of triangle $ABC$. Knowing that $\angle ACM = 30^o$ and $\angle AMB = 45^o$, find the measure of $\angle BAM$. p4. We denote as $S(X)$ the sum of the elements belonging to the set $X$. Divide the set $C = \{1,2,3,..., 2022\}$ into two disjoint subsets $A$ and $B$ so that |S(A)-S(B)| is minimum. PS. Problem 1 was also used as Seniors p1.
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Tags: chilean NMO, geometry, algebra, combinatorics, number theory
04.09.2022 19:12
04.09.2022 20:05
05.09.2022 04:37
parmenides51 wrote: p4. We denote as $S(X)$ the sum of the elements belonging to the set $X$. Divide the set $C = \{1,2,3,..., 2022\}$ into two disjoint subsets $A$ and $B$ so that |S(A)-S(B)| is minimum.
02.10.2022 19:43
I have solution for p1 If we try n=2 2n+1=5 4n+1=9 It not satisfying our problem But n=3 2n+1=7 4n+1=13 It is true statement And we know prime number =6k+-1 And we take n=6k+1 2n+1=12k+3=0 mod 3 it isnt satisfying And 6k-1=n 4n+1=24k-3=0 mod 3 it isnt satisfying our equation And we have only one answer it is n=3
27.06.2023 23:19
p1 First of all, let's prove that $n>3$ is not possible. Case 1.$n\equiv 1\pmod{3}$ $$2n+1\equiv 2\cdot 1+1\equiv 0\pmod{3}$$Which means that $3|2n+1$ which we cannot have,so there are no solutions in this case. Case 2.$ n\equiv 2\pmod{3}$ $$4n+1\equiv 4\cdot 2+1\equiv 9\equiv 0\pmod{3}$$Which means that $3|4n+1$ which we cannot have,so there are no solutions in this case either. Now we only need to check the cases when: $n=2\Rightarrow 4n+1=9\Rightarrow$No solutions. $n=3\Rightarrow 2n+1=7$, $4n+1=13$ $\Rightarrow n=3$ is a solution. Thus, we conclude that there exist only $1$ such number.