Question

Arătați ca 1+3^1+3^2+...+3^2003 se divide cu 455

Answer 1

1+3¹+3²+3³+3⁴+3⁵+.....+3²⁰⁰³=
=(1+3¹+3²)+3³(1+3¹+3²)+.....+3²⁰⁰⁶(1+3¹+3²)=
=(1+3+9)+3³(1+3+9)+.....+3²⁰⁰⁶(1+3+9)=
=13+3³ x13+.....+3²⁰⁰⁶x13=
=13(1+3³+.....+3²⁰⁰⁶)=
=13(1+3³+.....+3²⁰⁰⁶)=
=13[(1+3³)+3⁶(1+3³).....+3²⁰⁰³(1+3³)=
=13[(1+27)+3⁶(1+27).....+3²⁰⁰³(1+27)=
=13 x28(1+3⁶+.....+3²⁰⁰³)=
=13 x7x 4[(1+3⁶)+.....+3¹⁹⁹⁷(1+3⁶)]=
=13 x7x 4[(1+729)+.....+3¹⁹⁹⁷(1+729)]=
=13 x7x 4(730+.....+3¹⁹⁹⁷x730)=
=13 x7x 4x730(1+.....+3¹⁹⁹⁷)=
=13 x7x 4x5x146(1+.....+3¹⁹⁹⁷)=
=13 x7x 5x4x 146(1+.....+3¹⁹⁹⁷)=
=455x4x 146(1+.....+3¹⁹⁹⁷)=
deci divizibil cu 455