Question

Dem ca nr a=1+6+6^2+6^3+...+6^101 e divizibil cu 37x7x43

Answer 1

Termenii din suma pot fi grupati 2 cate 2
$1+6=7$
$6^{2}+6^{3}=6^{2}(1+6)=7*6^{2}$
$6^{4}+6^{5}=6^{4}(1+6)=7*6^{4}$
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$6^{100}+6^{101}=6^{100}(1+6)=7*6^{100}$
Acum le adunam pe toate
$a=1+6+6^{2}+6^{3}+..+6^{100}+6^{101}=7+7*6^{2}+7*6^{4}+..+7^{100}=7(1+6^{2}+6^{4}+..+6^{100})=7b$
notam acea suma cu b
Grupam din nou termenii 2 cate 2
$1+6^{2}=1+36=37$
$6^{4}+6^{6}=6^{4}(1+36)=37*6^{4}$
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$6^{98}+6^{100}=6^{98}(1+36)=37*6^{98}$
Adunam toti termenii
$b=1+6^{2}+6^{4}+6^{6}+6^{8}+..+6^{98}+6^{100}=37+37*6^{4}+..+37*6^{98}=37(1+6^{4}+..+6^{98})=37*c$
Am notat cu c suma ramasa
Acum grupam termenii 3 cate 3
$1+6^{4}+6^{8}=1680913$
$6^{12}+6^{16}+6^{20}=6^{12}(1+6^{4}+6^{8})=1680913*6^{12}$
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$6^{90}+6^{94}+6^{98}=6^{90}(1+6^{4}+6^{8})=1680913*6^{90}$

Atunci adunam toti termenii
$c=1+6+6^{4}+6^{8}+..+6^{98}=1680913+1680913*6^{10}+..+1680913*6^{90}=1680913(1+6^{12}+..+6^{90})=1680913d$
Numarul acela este divizibil cu 43
1680813=43*39091
Atunci
$c=43*39091*d$
Si atunci rezulta ca
$a=7*37*43*<span>39091*d</span>$ care e divizibil atunci cu 37*7*43