Question

va rog ,repede, dau coroana​

Answer 1

$\displaystyle\bf\\fie~n\in\mathbb{N},~astfel~incat~n:7=c_1,~rest~5,~si~n:6=c_2,~rest~4,~conform\\teoremei~impartirii~cu~rest~\implies n=7c_1+5,~si~n=6c_2+4,~inmultim~prima\\egalitate~cu~6~iar~cea~de~a~doua~o~inmultim~cu~7 \implies 6n=42c_1+30,~si~7n=42c_2+28,~efectuam~7n-6n=42(c_2-c_1)-2.$

$\displaystyle\bf\\si~cum~exista~o~infinitate~de~diferente~c_2-c_1 \implies \exists~o~infinitate~de~numere~n~\\cu~proprietatea~din~enunt.\\n=42(c_2-c_1)-2,~adunam~42~in~ambii~membri\implies n+42=42(c_2-c_1)+40,~si~cum~40<42~~rezulta~deci~ca~n+42~~da~restul~40~la~impartirea~cu~~42,~dar\\n+42~si~n~dau~acelasi~rest~prin~impartirea~la~42~\implies~n~da~restul~40\\la~impartirea~cu~42.$