1)
Observam ca [tex]6^{2018} = 6^2 \cdot 6^{2016} = 6^2 \cdot 6^{3\cdot 672}=36\cdot (6^{672})^3[/tex]. De asemenea, 36 poate fi scris astfel:
[tex]36=1 + 8 + 27 =1^3 + 2^3 + 3^3[/tex]
Acum, rescriem:
[tex]6^{2018}=36\cdot (6^{672})^3=(1^3 + 2^3 + 3^3)(6^{672})^3[/tex]
[tex]6^{2018}=(6^{672})^3 + 2^3\cdot (6^{672})^3 + 3^3 \cdot (6^{672})^3[/tex]
[tex]6^{2018}=(6^{672})^3 + (2 \cdot 6^{672})^3 + (3 \cdot 6^{672})^3[/tex]
Astfel, am scris [tex]6^{2018}[/tex] ca suma de 3 cuburi perfecte.
2)
Avem:
[tex]\overline{ab} + \overline{bba} + 10b=10a+b+100b+10b+a+10b=121b+11a[/tex]
si
[tex]\overline{ba}+\overline{aab}+10a=10b+a+100a+10a+b+10a=121a+11b[/tex]
Deci, putem scrie numerele astfel:
[tex]\overline{ab}+\overline{bba}+10b=11(11b+a)[/tex]
[tex]\overline{ba}+\overline{aab}+10a=11(11a+b)[/tex]
Asadar, numerele date sunt divizibile cu 11.
