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Question Number 119956 by bobhans last updated on 28/Oct/20
Given a = 1+3+3^2 +3^3 +3^4 +...+3^(100) Find the remainder of dividing the number by 5 . (a) 2 (b) 0 (c)4 (d)1 (e) 3
$${Given}\:{a}\:=\:\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +\mathrm{3}^{\mathrm{4}} +…+\mathrm{3}^{\mathrm{100}} \\ $$$${Find}\:{the}\:{remainder}\:{of}\:{dividing}\:{the}\:{number} \\ $$$${by}\:\mathrm{5}\:. \\ $$$$\left({a}\right)\:\mathrm{2}\:\:\:\:\:\left({b}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({c}\right)\mathrm{4}\:\:\:\:\:\:\left({d}\right)\mathrm{1}\:\:\:\:\:\:\left({e}\right)\:\mathrm{3} \\ $$
Answered by talminator2856791 last updated on 08/Sep/21
a = Σ_(k = 0) ^(24) 3^(4k) (1+3^2 ) + Σ_(k = 0) ^(24) 3^(4k+1) (1+3^2 ) + 3^(100) Σ_(k = 0) ^(24) 3^(4k) (1+3^2 ) = 10Σ_(k = 0) ^(24) 3^(4k) Σ_(k = 0) ^(24) 3^(4k) (1+3^2 ) = 10Σ_(k = 0) ^(24) 3^(4k+1) → 5 ∣10Σ_(k = 0) ^(24) 3^(4k) , 5∣10Σ_(k = 0) ^(24) 3^(4k+1) → 5∣1+3+3^2 +.....+3^(99) ⇒ 5∣3(1+3+3^2 +.....+3^(99) ) = 5∣3+3^2 +3^3 +.....+3^(100) ⇒ 1+3+3^2 +......+3^(100) ≡ 1 (mod 5) answer: d
$$\:{a}\:=\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}} \left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)\:+\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\mathrm{3}^{\mathrm{4}{k}+\mathrm{1}} \left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)\:+\:\mathrm{3}^{\mathrm{100}} \: \\ $$$$\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}} \left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)\:=\:\mathrm{10}\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}} \\ $$$$\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}} \left(\mathrm{1}+\mathrm{3}^{\mathrm{2}} \right)\:=\:\mathrm{10}\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}+\mathrm{1}} \\ $$$$\:\rightarrow\:\mathrm{5}\:\mid\mathrm{10}\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}} \:,\:\:\:\mathrm{5}\mid\mathrm{10}\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{24}} {\sum}}\:\mathrm{3}^{\mathrm{4}{k}+\mathrm{1}} \\ $$$$\:\rightarrow\:\mathrm{5}\mid\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +…..+\mathrm{3}^{\mathrm{99}} \\ $$$$\: \\ $$$$\:\Rightarrow\:\mathrm{5}\mid\mathrm{3}\left(\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +…..+\mathrm{3}^{\mathrm{99}} \right) \\ $$$$\:=\:\:\mathrm{5}\mid\mathrm{3}+\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{3}} +…..+\mathrm{3}^{\mathrm{100}} \\ $$$$\: \\ $$$$\:\Rightarrow\:\mathrm{1}+\mathrm{3}+\mathrm{3}^{\mathrm{2}} +……+\mathrm{3}^{\mathrm{100}} \:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right) \\ $$$$\:\mathrm{answer}:\:{d} \\ $$
Answered by TANMAY PANACEA last updated on 28/Oct/20
S=((3^(101) −1)/(3−1))=(1/2)(3^(101) −1) 3^1 =3 3^2 =9 3^3 =27 3^4 =81 3^5 =243 3^6 =729 so last digit repets (3→9→7→1) ((101)/4)→25 cycle of(3→9→7→1)+plus 1 last digit of 3^(101) is 3 (now last digit 3)−1 so last digit becomes=2 when devided by 2 last digit becomes 1 ((3^(101) −1)/2)= last digit=1 ((3^(101) −1)/2)=(5×even multiple )+last digit=1
$${S}=\frac{\mathrm{3}^{\mathrm{101}} −\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}^{\mathrm{101}} −\mathrm{1}\right) \\ $$$$\mathrm{3}^{\mathrm{1}} =\mathrm{3} \\ $$$$\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{3}^{\mathrm{3}} =\mathrm{27} \\ $$$$\mathrm{3}^{\mathrm{4}} =\mathrm{81} \\ $$$$\mathrm{3}^{\mathrm{5}} =\mathrm{243} \\ $$$$\mathrm{3}^{\mathrm{6}} =\mathrm{729} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{repets}}\:\left(\mathrm{3}\rightarrow\mathrm{9}\rightarrow\mathrm{7}\rightarrow\mathrm{1}\right) \\ $$$$\frac{\mathrm{101}}{\mathrm{4}}\rightarrow\mathrm{25}\:{cycle}\:{of}\left(\mathrm{3}\rightarrow\mathrm{9}\rightarrow\mathrm{7}\rightarrow\mathrm{1}\right)+\boldsymbol{{plus}}\:\mathrm{1} \\ $$$$\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{of}}\:\mathrm{3}^{\mathrm{101}} \:\boldsymbol{{is}}\:\:\mathrm{3} \\ $$$$\left(\boldsymbol{{now}}\:\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\mathrm{3}\right)−\mathrm{1} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{becomes}}=\mathrm{2} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{devided}}\:\boldsymbol{{by}}\:\mathrm{2} \\ $$$$\boldsymbol{{last}}\:\boldsymbol{{digit}}\:\boldsymbol{{becomes}}\:\mathrm{1} \\ $$$$\frac{\mathrm{3}^{\mathrm{101}} −\mathrm{1}}{\mathrm{2}}=\:\:{last}\:{digit}=\mathrm{1} \\ $$$$\frac{\mathrm{3}^{\mathrm{101}} −\mathrm{1}}{\mathrm{2}}=\left(\mathrm{5}×{even}\:{multiple}\:\right)+\boldsymbol{{last}}\:\boldsymbol{{digit}}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

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