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Question Number 91448 by Cynosure last updated on 30/Apr/20

prove that 1+x^(111) +x^(222) +x^(333) +x^(444)   divides 1+ x^(111) +x^(222) +x^(333) +.......+x^(999)

$${prove}\:{that}\:\mathrm{1}+{x}^{\mathrm{111}} +{x}^{\mathrm{222}} +{x}^{\mathrm{333}} +{x}^{\mathrm{444}} \:\:{divides}\:\mathrm{1}+\:{x}^{\mathrm{111}} +{x}^{\mathrm{222}} +{x}^{\mathrm{333}} +.......+{x}^{\mathrm{999}} \\ $$

Commented by Cynosure last updated on 30/Apr/20

pls help me on this

$${pls}\:{help}\:{me}\:{on}\:{this}\: \\ $$

Commented by Prithwish Sen 1 last updated on 30/Apr/20

putting x^(111)  = a  then ((1+a+a^2 +.....+a^9 )/(1+a+a^2 +a^3 +a^4 )) = ((1−a^(10) )/(1−a^5 )) = ((1−(a^5 )^2 )/(1−a^5 )) = 1+a^5    = 1+x^(555) proved.

$$\mathrm{putting}\:\mathrm{x}^{\mathrm{111}} \:=\:\mathrm{a} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +.....+\mathrm{a}^{\mathrm{9}} }{\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +\mathrm{a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{4}} }\:=\:\frac{\mathrm{1}−\mathrm{a}^{\mathrm{10}} }{\mathrm{1}−\mathrm{a}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}−\left(\mathrm{a}^{\mathrm{5}} \right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{a}^{\mathrm{5}} }\:=\:\mathrm{1}+\mathrm{a}^{\mathrm{5}} \\ $$$$\:=\:\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{555}} \boldsymbol{\mathrm{proved}}. \\ $$

Commented by Cynosure last updated on 30/Apr/20

    i think it should be solved in this manner  then ((1+a+a^2 +.....+a^9 )/(1+a+a^2 +a^3 +a^4 )) = ((a^(10) −1)/(a^5 −1)) = (((a^5 )^2 −1)/(a^5 −1)) = a^5 +1

$$ \\ $$$$ \\ $$$${i}\:{think}\:{it}\:{should}\:{be}\:{solved}\:{in}\:{this}\:{manner} \\ $$$$\mathrm{then}\:\frac{\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +.....+\mathrm{a}^{\mathrm{9}} }{\mathrm{1}+\mathrm{a}+\mathrm{a}^{\mathrm{2}} +\mathrm{a}^{\mathrm{3}} +\mathrm{a}^{\mathrm{4}} }\:=\:\frac{\mathrm{a}^{\mathrm{10}} −\mathrm{1}}{\mathrm{a}^{\mathrm{5}} −\mathrm{1}}\:=\:\frac{\left(\mathrm{a}^{\mathrm{5}} \right)^{\mathrm{2}} −\mathrm{1}}{{a}^{\mathrm{5}} −\mathrm{1}}\:=\:\mathrm{a}^{\mathrm{5}} +\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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