14310

14310 (Traian Covaciu)

Fie $n,n+2,n+6$ trei numere naturale și $S$ suma lor.

a) Dați exemplu de cel puțin trei valori pentru $n\in \mathbb {N}$ astfel incat numerele $n,n+2,n+6$ sa fie simultan numere prime.

b) Daca Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n, n + 2, n + 6 } sunt simultan numere prime, aratati ca exista Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N} } astfel incat Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S = 9k + 5} .

c) Daca Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n, n + 2, n + 6 } sunt numere prime, determinati restul impartirii numarului Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} la Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 18 } .

Solutie:

a) Pentru Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n = 5 } numerele sunt $5,7,11$. Pentru $n=11$ avem $11,13,17$, iar pentru $n=17$ obtinem $17,19,23$.

b) Se stie ca numerele prime au forma $6p+1$ sau $6p+5$. Daca $n=6p+1$, atunci $n+2=6p+3$ care nu este numar prim. Asadar, $n=6p+5$. In aceasta situatie avem $S=6p+5+6p+7+6p+11=18p+23=9(2p+2)+5$. In concluzie, exista $k=2p+2$ astfel incat $S=9k+5$.

c) Din punctul b) avem $S=18(p+1)+5$, deci restul impartirii lui $S$ la $18$ este $5$.