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Question Number 25046 by Tinkutara last updated on 02/Dec/17

Show that (a) N=((10^(143) −1)/9) is composite, and (b) N has two factors each of which is a series of a G.P.

$${Show}\:{that} \\ $$$$\left({a}\right)\:{N}=\frac{\mathrm{10}^{\mathrm{143}} −\mathrm{1}}{\mathrm{9}}\:{is}\:{composite},\:{and} \\ $$$$\left({b}\right)\:{N}\:{has}\:{two}\:{factors}\:{each}\:{of}\:{which}\:{is} \\ $$$${a}\:{series}\:{of}\:{a}\:{G}.{P}. \\ $$

Answered by Rasheed.Sindhi last updated on 05/Dec/17

N(10)=((10^(143) −1)/9)=((10^(143) −1)/(10−1)) In general N(x)=((x^(143) −1)/(x−1)) =(((x−1)(x^(142) +x^(141) +x^(140) +...+x+1))/((x−1))) =x^(142) +x^(141) +x^(140) +...+x+1 So, N(10)=10^(142) +10^(141) +...+10+1 [N(10)=111...11 (143 times) 143=11×13 So the number can be grouped either 11 groups of 13 ′ones′ or 13 groups of 11 ′ones′.] =(10^(12) +10^(11) +...10+1).10^(130) +(10^(12) +10^(11) +...10+1).10^(117) +(10^(12) +10^(11) +...10+1).10^(104) +...+(10^(12) +10^(11) +...10+1).10^(13) +(10^(12) +10^(11) +...10+1).1 =(10^(12) +10^(11) +..10+1)(10^(130) +10^(117) +...10^(13) +1) (a)Hence the number is composite. (b)Both factors are geometric series. 10^(12) +10^(11) +10^(10) ...+10+1:common ratio 10^(−1) . 10^(130) +10^(117) +10^(104) +...+10^(13) +1: common ratio 10^(−13)

$${N}\left(\mathrm{10}\right)=\frac{\mathrm{10}^{\mathrm{143}} −\mathrm{1}}{\mathrm{9}}=\frac{\mathrm{10}^{\mathrm{143}} −\mathrm{1}}{\mathrm{10}−\mathrm{1}} \\ $$$$\:\:\mathrm{In}\:\mathrm{general} \\ $$$$\:\:\:{N}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{143}} −\mathrm{1}}{\mathrm{x}−\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{142}} +\mathrm{x}^{\mathrm{141}} +\mathrm{x}^{\mathrm{140}} +...+\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{x}−\mathrm{1}\right)} \\ $$$$\:\:\:\:=\mathrm{x}^{\mathrm{142}} +\mathrm{x}^{\mathrm{141}} +\mathrm{x}^{\mathrm{140}} +...+\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{So}, \\ $$$${N}\left(\mathrm{10}\right)=\mathrm{10}^{\mathrm{142}} +\mathrm{10}^{\mathrm{141}} +...+\mathrm{10}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{N}\left(\mathrm{10}\right)=\mathrm{111}...\mathrm{11}\:\left(\mathrm{143}\:\mathrm{times}\right)\right. \\ $$$$\:\:\:\:\:\mathrm{143}=\mathrm{11}×\mathrm{13} \\ $$$$\:\:\:\mathrm{So}\:\mathrm{the}\:\mathrm{number}\:\mathrm{can}\:\mathrm{be}\:\mathrm{grouped}\:\mathrm{either} \\ $$$$\left.\:\:\:\mathrm{11}\:\mathrm{groups}\:\mathrm{of}\:\mathrm{13}\:'\mathrm{ones}'\:\mathrm{or}\:\mathrm{13}\:\mathrm{groups}\:\mathrm{of}\:\mathrm{11}\:'\mathrm{ones}'.\right] \\ $$$$\:\:\:=\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +...\mathrm{10}+\mathrm{1}\right).\mathrm{10}^{\mathrm{130}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +...\mathrm{10}+\mathrm{1}\right).\mathrm{10}^{\mathrm{117}} \\ $$$$\:+\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +...\mathrm{10}+\mathrm{1}\right).\mathrm{10}^{\mathrm{104}} \\ $$$$\:\:+...+\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +...\mathrm{10}+\mathrm{1}\right).\mathrm{10}^{\mathrm{13}} \\ $$$$\:+\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +...\mathrm{10}+\mathrm{1}\right).\mathrm{1} \\ $$$$ \\ $$$$=\left(\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +..\mathrm{10}+\mathrm{1}\right)\left(\mathrm{10}^{\mathrm{130}} +\mathrm{10}^{\mathrm{117}} +...\mathrm{10}^{\mathrm{13}} +\mathrm{1}\right) \\ $$$$\left(\mathrm{a}\right)\mathrm{Hence}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{composite}. \\ $$$$\left(\mathrm{b}\right)\mathrm{Both}\:\mathrm{factors}\:\mathrm{are}\:\mathrm{geometric}\:\mathrm{series}. \\ $$$$\mathrm{10}^{\mathrm{12}} +\mathrm{10}^{\mathrm{11}} +\mathrm{10}^{\mathrm{10}} ...+\mathrm{10}+\mathrm{1}:\mathrm{common}\:\mathrm{ratio}\:\mathrm{10}^{−\mathrm{1}} . \\ $$$$\mathrm{10}^{\mathrm{130}} +\mathrm{10}^{\mathrm{117}} +\mathrm{10}^{\mathrm{104}} +...+\mathrm{10}^{\mathrm{13}} +\mathrm{1}:\:\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{10}^{−\mathrm{13}} \\ $$

Commented by Tinkutara last updated on 05/Dec/17

Thank you Sir! There is one more answer: N=(1+10^(11) +10^(22) +...+10^(132) )(1+10+ 10^2 +...+10^(10) )

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$$${There}\:{is}\:{one}\:{more}\:{answer}: \\ $$$${N}=\left(\mathrm{1}+\mathrm{10}^{\mathrm{11}} +\mathrm{10}^{\mathrm{22}} +...+\mathrm{10}^{\mathrm{132}} \right)\left(\mathrm{1}+\mathrm{10}+\right. \\ $$$$\left.\mathrm{10}^{\mathrm{2}} +...+\mathrm{10}^{\mathrm{10}} \right) \\ $$

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