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1991 Chile Classification NMO

Part of Chile Classification NMO

Subcontests

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1991 Chile Classification / Qualifying NMO III

p1. One hundred elements can be chosen from {1,2,...,200}\{1,2,...,200\}, so that no one element chosen is divisor of another chosen element. Give an example of this choice, and prove that this is impossible for 101101 elements.
p2. Prove that the equation x+1990y=z2x + 1990y = z^2 has an infinite number of natural solutions.
p3. Find the quantity of natural numbers, such that none of their digits is equal to 11, and that the product of its digits is equal to 4848.
p4. Given a sheet of material flexible, as in the first figure, and gluing the edges that have the same letters as indicated by the cast, we get a bull. What is obtained by the same procedure, starting from the second figure? https://cdn.artofproblemsolving.com/attachments/d/f/66d4242403a1d917352c58dafe4d95794bd52f.png
p5. The sequence (Cn)(C_n), n>0n> 0 of integers is defined by the relations: ∙\bullet C0=0C_0 = 0 ∙\bullet C2n=CnC_{2n} = C_n ∙\bullet C2n+1=1−CnC_{2n + 1} = 1-C_n Determine C1991C_{1991}.
p6. Find a finite sequence C0,C1,...,C1991C_0, C_1,..., C_{1991} of naturals, check the following condition for everything a∈{0,1,...,1991}a\in \{0,1,..., 1991\}, CaC_a is the number of numbers aa in the sequence.
p7. In a △ABC\vartriangle ABC, with orthocenter HH, let A′A', B′B', C′C' be the midpoints of the sides, and A′′A'', B′′B'', C′′C'' be the midpoints of the segments that join the orthocenter with the vertices. Prove that points A′A', B′B', C′C', A′′A'', B′′B'', C′′C'' are concyclic.