Question

$Sa~se~arate~ca~daca~numarul~\underbrace{111...1}~este~prim,~atunci~n~este~prim. \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~n ~cifre\\ Reciproca~este~adevarata~? \\ \\ ------------------------------- \\ \\ Problema~a~aparut~in~Gazeta~Matematica,~numarul~5-6/1988~la~ \\ \\ sectiunea~"PROBLEME~PRELIMINARE~DE~TIP~O.I.M.~ \\ \\ PENTRU~ELEVII~DE~GIMNAZIU".$

$Abordarea~mea~a~fost~urmatoarea: \\ \\ Am~inceput~cu~sfarsitul,~adica~cu~intrebarea~"Reciproca~este~ \\ \\ adevarata?".~Evindent,~nu!~n=3~este~prim,~insa~111~nu~este~prim! \\ \\ Pentru~prima~parte~a~problemei~m-am~gandit~ca~daca~111...1~este \\ \\ prim~atunci~criteriul~de~divizibilitate~cu~11~nu~este~valabil,~si~prin~ \\ \\ urmare~rezulta~ca~n~este~impar.~Se~mai~pot~eventual~exclude \\ \\ multiplii~lui~3...insa~nu~am~idee~cum~s-ar~finaliza.$

Answer 1

[tex]\underbrace{1111...1}_{\mbox{n\ cifre}}= \frac{999.9(de \ n\ ori)}{9} = \frac{10^n-1}{9} prim\\ Presupunem \ ca \ n\ nu \ este \ prim=\ \textgreater \ n=kp\\ \frac{10^{kp}-1}{9} =\frac{(10^k)^{p}-1^p}{9} = \frac{(10^k-1)t}{9} = \frac{999...9 (de k ori)\cdot t}{9}=\\ =111...1(dekori)\cdot t=\ \textgreater \ \underbrace{1111...1}_{\mbox{n\ cifre}} \ nu\ este\ prim(Fals)\\ =\ \textgreater \ presupunerea\ facuta\ este\ falsa\ =\ \textgreater \ n\ prim.[/tex]

Answer 2

PP RA n=pm (n nu este prim)
[tex]\underbrace{111...1}={\underbrace{111...1}\underbrace{111...1}...\underbrace{111...1}}=\\ ~~~pm~~~~~~\underbrace{~~~~p~~~~~~~p~~~~~~~~~~~p}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~m\\ =\underbrace{111...1}\cdot\underbrace{100...0100...0\ ...\ 100...0}\ \vdots\underbrace{111...1}\\ ~~~~~~~~p~~~~~~~de~m~ori~1~si~de~\\ ~~~~~~~~~~~~~~~~~~(m-1)p~ori~0 [/tex]
Am obtinut o condradictie. Deci n este prim.