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Question Number 34159 by candre last updated on 01/May/18

$$letS=1+2+...+2018$$$$computeS\left(\mathrm{mod}2\right)+S\left(\mathrm{mod}8\right)+S\left(\mathrm{mod}2018\right)$$

Commented by Rasheed.Sindhi last updated on 05/May/18

$$\mathrm{S}=(1+3+5+...+2017)+(2+4+...+2018)$$$$\mathrm{Sum}\mathrm{of}\mathrm{even}\mathrm{numbers}\mathrm{is}\mathrm{even}\mathrm{number}$$$$\therefore 2+4+...+2018\mathrm{even}\mathrm{number}$$$$\mathrm{Sum}\mathrm{of}\mathrm{odd}\mathrm{numbers},\mathrm{when}\mathrm{number}\mathrm{of}$$$$\mathrm{numbers}\mathrm{is}\mathrm{odd},\mathrm{is}\mathrm{odd}.$$$$\therefore 1+3+5+...+2017\mathrm{is}\mathrm{odd}\mathrm{number}$$$$\therefore \mathrm{S}=\mathrm{odd}+\mathrm{even}=\mathrm{odd}\mathrm{number}$$$$\therefore \mathrm{S}\left(\mathrm{mod}2\right)=1$$$$-----$$$$\mathrm{S}\left(\mathrm{mod}8\right)?$$$$------$$$$\mathrm{S}=(1+2+3+...+2017)+2018$$$$=(1+2017)+(2+2016)+...+1009+2018$$$$=2018+2018+...+1009+2018$$$$=\left(\mathrm{multiple}\mathrm{of}2018\right)+1009$$$$\therefore \mathrm{S}\left(\mathrm{mod}2018\right)=1009$$$$\mathrm{Continue}$$

Answered by Rasheed.Sindhi last updated on 05/May/18

$$\mathrm{S}=1+2+...+2018;\mathrm{an}\mathrm{AP}:a=1,\mathrm{d}=1,\mathrm{n}=2018$$$$\mathrm{S}=\frac{n}{2}[2a+(n-1)d]$$$$\mathrm{S}=\frac{2018}{2}[2\left(1\right)+(2018-1)\left(1\right)]$$$$=1009[2018+1]$$$$=2018\times 1009+1009$$$$\therefore \mathrm{S}\left(\mathrm{mod}2018\right)=1009.........\mathrm{A}$$$$-----------$$$$\mathrm{S}=1009\times 2019$$$$=(1008+1)(2018+1)$$$$=1008\times 2018+1008+2018+1$$$$=1008\times 2019+2018+1$$$$=2(504\times 2019+1009)+1$$$$\therefore \mathrm{S}\left(\mathrm{mod}2\right)=1................\mathrm{B}$$$$------$$$$\mathrm{S}=1009\times 2019$$$$=\left(1009\right)(2000+16+3)$$$$=2000\times 1009+16\times 1009+3\times 1009$$$$=2000\times 1009+16\times 1009+3000+27$$$$=2000\times 1009+16\times 1009+3000+24+3$$$$2000,16,3000\mathrm{\&}24\mathrm{are}\mathrm{divisible}\mathrm{by}8$$$$[\mathrm{Complete}\mathrm{thousands}\mathrm{are}\mathrm{always}\mathrm{divisible}$$$$\mathrm{by}8]$$$$\therefore \mathrm{S}\left(\mathrm{mod}8\right)=3...............\mathrm{C}$$$$\mathrm{From}\mathrm{A},\mathrm{B}\mathrm{\&}\mathrm{C}$$$$\mathrm{S}\left(\mathrm{mod}2\right)+\mathrm{S}\left(\mathrm{mod}8\right)+\mathrm{S}\left(\mathrm{mod}2018\right)$$$$=1+3+1009=1013$$

Commented by Rasheed.Sindhi last updated on 09/May/18

$$\mathrm{Mr}.\mathrm{candre}\mathrm{please}\mathrm{confirm}\mathrm{the}\mathrm{answer}.$$

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