Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 154401 by mathdanisur last updated on 18/Sep/21

$$\mathrm{prove}\mathrm{that}\mathrm{the}\mathrm{sum}$$$$2+2^2+2^3+2^4+...+2^{2019}+2^{2020}$$$$\mathrm{is}\mathrm{divisible}\mathrm{by}30$$

Answered by talminator2856791 last updated on 18/Sep/21

$$$$$$2^{k+1}+2^{k+3}=2^k(8+2)$$$$=10\cdot 2^k$$$$$$$$2^{k+2}+2^{k+4}=2^k(16+4)$$$$=20\cdot 2^k$$$$$$$$2^{k+1}+2^{k+2}+2^{k+3}+2^{k+4}=30\cdot 2^k$$$$$$$$S=2+2^2+2^3+2^4+.....+2^{2019}+2^{2020}$$$$=(2+2^2+2^3+2^4)+(2^5+2^6+2^7+2^8)+.....$$$$+(2^{2017}+2^{2018}+2^{2019}+2^{2020})$$$$=(2+2^2+2^3+2^4)+2^4(2+2^2+2^3+2^4)+.....$$$$+2^{2016}(2+2^2+2^3+2^4)$$$$=30+2^4\cdot 30+2^8\cdot 30+.....+2^{2016}\cdot 30$$$$=30(1+2^4+.....+2^{2016})$$$$$$$$\Rightarrow 30\mid S$$$$$$

Commented by mathdanisur last updated on 18/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser}\mathrm{thanks}$$

Answered by Rasheed.Sindhi last updated on 18/Sep/21

$$\underset{―}{\mathbb{MOD}30}$$$$2^4\equiv 16\Rightarrow 2^{4\times 1}\equiv 16$$$${\left(2^4\right)}^2\equiv {\left(16\right)}^2\Rightarrow 2^8\equiv 16\Rightarrow 2^{4\times 2}\equiv 16$$$${\left(2^8\right)}^2\equiv {\left(16\right)}^2\Rightarrow 2^{12}\equiv 16\Rightarrow 2^{4\times 3}\equiv 16$$$$\dots \dots \dots \dots \dots$$$$2^{4\times \mathrm{k}}\equiv 16$$$$2^{4\mathrm{k}}.2\equiv 16.2\equiv 2\Rightarrow 2^{4\mathrm{k}+1}\equiv 2$$$$2^{4\mathrm{k}+1}.2\equiv 2.2\equiv 4\Rightarrow 2^{4\mathrm{k}+2}\equiv 4$$$$2^{4\mathrm{k}+2}.2\equiv 4.2\equiv 8\Rightarrow 2^{4\mathrm{k}+3}\equiv 8$$$$2^{4\mathrm{k}}\equiv 16\Rightarrow 2^{4\mathrm{k}}\equiv 16$$$$▸\mathbb{S}=2+2^2+2^3+2^4+...+2^{2019}+2^{2020}$$$$=\underset{\mathrm{k}=0}{\stackrel{505}{\mathrm{\Sigma }}}(2^{4\mathrm{k}+1}+2^{4\mathrm{k}+2}+2^{4\mathrm{k}+3}+2^{4\mathrm{k}})$$$$\equiv \underset{\mathrm{k}=1}{\stackrel{505}{\mathrm{\Sigma }}}(2+4+8+16).\mathrm{k}=\underset{\mathrm{k}=1}{\stackrel{505}{\mathrm{\Sigma }}}\left(30\mathrm{k}\right)$$$$\mathbb{S}=30\times \underset{\mathrm{k}=1}{\stackrel{505}{\mathrm{\Sigma }}}\mathrm{k}$$$$\therefore 30\mid \mathbb{S}$$

Commented by mathdanisur last updated on 18/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser}\mathrm{thankyou}$$

Answered by JDamian last updated on 18/Sep/21

$$S=2\frac{2^{2020}-1}{2-1}=2(2^{2020}-1)$$$$S=2^{2021}-2$$$$Smod30=(2^{2021}-2)mod30=$$$$=2^{2021}mod30-2$$$$$$$$\mid {\left(2^5\right)}^{k}mod30={(30+2)}^{k}mod30=$$$$=2^{k}mod30\mid$$$$$$$$2^{2021}mod30=[2\cdot {\left(2^5\right)}^{404}]mod30=$$$$=(2\cdot 2^{404})mod30=2^{405}mod30=$$$$={\left(2^5\right)}^{81}mod30=2^{81}mod30=$$$$=2\cdot {\left(2^5\right)}^{16}mod30=2^{17}mod30=$$$$=2^2{\left(2^5\right)}^{3}mod30=2^{5}mod30=2$$$$$$$$Smod30=2-2=0$$

Commented by mathdanisur last updated on 18/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser}\mathrm{thankyou}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com