Question

determinati restul impartirii numarului 3^2005+3^2006+3^2007 la 244

Answer 1

$\displaystyle Folosim~urmatoarea~proprietate: \\ \\ (a+b)^n=M_a+b^n,~unde~a,b \in \mathbb{Z},~iar~n \in \mathbb{N^*},~iar~M_a~ \\ \\ desemnseaza~un~multiplu~arbitrar~de-al~lui~a. \\ \\ 3^{2015}=(3^5)^{403}=243^{403}=(244-1)^{403}=M_{244}+(-1)^{403}= \\ \\ =M_{244}-1. \\ \\ 3^{2016}=3 \cdot 3^{2015}=3(M_{244}-1)=3M_{244}-3=M_{244}-3. \\ \\ 3^{2017}=9 \cdot 3^{2015}=9(M_{244}-1)=9M_{244}-9=M_{244}-9. \\ \\ 3^{2015}+2^{2016}+3^{2017}=M_{244}-1+M_{244}-3+M_{244}-9=$

$\displaystyle =M_{244}-13=M_{244}+244-13=M_{244}+231 \Rightarrow~restul=231. \\ \\ *Asa~cum~spuneam~la~inceput,~notatia~M_{244}~desemneaza~un \\ \\ multiplu~arbitrar~a~lui~244~si~nu~neaparat~unul~si~acelasi~numar.$