Question

sa se arate ca (7^15-2^39):5 (3^62+4^58):5.

Answer 1

   
[tex]\text{Vom arata ca: } ~~ (7^{15}-2^{39})~\vdots~5 \\ \text{Pentru ca expresia din paranteza sa fie divizibila cu 5, } \\ \text{trebuie sa aiba ultima cifra 0 sau 5. } \\ Calculam ~ultima ~cifra~(U) ~a ~expresiei:~~~7^{15}-2^{39} \\ U(7^{15}-2^{39}) = U(7^{15})-U(2^{39}) = U(7^{12+3})-U(2^{36+3})= \\ =U(7^{12}\times7^{3})-U(2^{36}\times 2^3)=U(7^{4 \times 3}\times7^{3})-U(2^{4 \times 9}\times 2^3)= \\ =U((7^4)^3 \times7^{3})-U((2^4)^9\times 2^3)=/tex]

[tex] =U(2401^3 \times 343)-U(16^9\times 8)= \\ =U(343-U(6\times 8)=U(343-48)=\boxed{5} \\ \Longrightarrow~~\boxed{(7^{15}-2^{39})~ \vdots~ 5}[/tex]



[tex]\text{Vom arata ca: } ~~ (3^{62}+4^{58})~\vdots~5 \\ U(3^{62}+4^{58}) = U(3^{60+2}+4^{58})=U(3^{60} \times 3^2}+4^{58})= \\ =U(3^{4 \times 15} \times 9}+4^{2 \times 29})= U(81^{15} \times 9}+16^{29})= \\ =U(1 \times 9 + 6) = U(9+6) = \boxed{5} \\ \Longrightarrow ~~ \boxed{ (3^{62}+4^{58})~\vdots~5 }[/tex]