Question

Aratati că numărul a egal 1 plus 6 plus 6 la a doua plus puncte puncte plus 6 la 101 este divizibil cu 7 ori 37 ori 43 ​

Answer 1

 

$a=6^0+6^1+6^2+6^3+6^4+6^5+\cdots+6^{99}+6^{100}+6^{101}\\\text{Caolculam numarul de termeini. }~n=?\\\text{Exponentii sunt de la 0 la 101.}\\\Implies~n=101+1=\boxed{\bf102~termeni}$

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$\displaystyle\\ \boxed{\bf7}\\\text{Observam ca }~~6^0+6^1 = 1 + 6 = 7\\\\\text{Vom grupa termenii in grupe de cate 2 termeni.}\\\text{Verificam daca avem voie.}\\\boxed{\bf~n=102~si~102~\vdots~2}\\\\(6^0+6^1)+(6^2+6^3)+(6^4+6^5)+\cdots+(6^{100}+6^{101})=\\\\=(1+6)+6^2(1+6)+6^4(1+6)+\cdots+6^100(1+6})=\\\\=7+6^2\times7+6^4\times7+\cdots+6^{100}\times7=\\\\=\boxed{\bf7(1+6^2+6^4+\cdots+6^{100})~\vdots~7}$

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$\displaystyle\\\boxed{\bf43}\\\text{Observam ca }~~6^0+6^1 +6^2= 1 + 6 +36 = 43\\\\\text{Vom grupa termenii in grupe de cate 3 termeni.}\\\text{Verificam daca avem voie.}\\\boxed{\bf~n=102~si~102~\vdots~3}\\\\(6^0+6^1+6^2)+(6^3+6^4+6^5)+\cdots+(6^{99}+6^{100}+6^{101})=\\\\=(1+6+6^2)+6^3(1+6+6^2)+\cdots+6^{99}(1+6+6^2)=\\\\=43+6^3\times43+\cdots+6^{99}\times43=\\\\=\boxed{\bf43(1+6^3+\cdots+6^{99})~\vdots~43}$

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$\displaystyle\\\boxed{\bf37}\\\text{Observam ca }~~6^0+6^1 +6^2+6^3=1+6+36+216=259=7\times37~\vdots~37\\\\\text{Vom grupa termenii in grupe de cate 4 termeni.}\\\text{Verificam daca avem voie.}\\n=102~si~102~\text{NU se divide cu 4}\\\\\implies~~a=6^0+6^1+6^2+\cdots+6^{101}~~~\text{Nu se divide cu 37.}\\\\\implies~~a=6^0+6^1+6^2+\cdots+6^{101}~~~\text{Nu este divizibil cu }~7\times37\times43.}\\\\\text{Daca aveam 107 in loc de 101 atuci n = 108 divizibil cu 2, cu 3 si cu 4.}$

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