Question

Determinați un număr natural n pentru care numerele n+1, n+2, n+3, ... , n+100 sunt numere compuse.​

Answer 1

$\displaystyle\bf\\\boxed{\bf un~numar~este~compus~cand~are~numarul~divizorilor~mai~mare~decat~2.}\\\\n+1,n+2,n+3,...,n+100~sunt~compuse.\\ca~sa~fim~siguri~ca~am~ales~corect,~si~numarul~nu~este~minim~sau~\\maxim~\implies ~folosim~ca~k!=1\cdot2\cdot...\cdot k,~este~compus~pentru~n\neq 2.\\dar,~1|k!,~2|k!,...100|k!,~de~unde~rezulta~ca~putem~alege~k=100,~pentru~ca\\100!=1\cdot 2\cdot 3\cdot...\cdot 100~care~respecta~conditia~de~mai~sus.$

$\displaystyle\bf\\deci,~\boxed{\bf n=100!},~desigur,~n~poate~fi~si~101!,~102!,.....~etc.$