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Question Number 126682 by bramlexs22 last updated on 23/Dec/20

$$Findtheremainder$$$$7+77+777+7777+...+\underset{2020times}{\underbrace{777...7}}$$$$whendividedby9.$$

Answered by talminator2856791 last updated on 23/Dec/20

$$$$$$7\underset{k=1}{\sum ^{2020}}k\left({10}^{2020-k}\right)=7\underset{k=1}{\sum ^{2020}}\underset{i=1}{\sum ^k}{10}^{i-1}$$$$$$$${10}^n\equiv 1\mathrm{mod}9,n\in \mathbb{N}$$$$$$$$\Rightarrow \frac{7\underset{k=1}{\sum ^{2020}}\underset{i=1}{\sum ^k}{10}^{i-1}}{9}=9j+7\left(2020\right)+7\underset{k=1}{\sum ^{2019}}k$$$$=9j+14140+7\left(2019\right)\left(1010\right)$$$$=9j+14288470$$$$$$$$\frac{14288470}{9}=1587607.7777.....$$$$$$$$\Rightarrow 7+77+777+.......+\underset{2020\mathrm{times}}{\underbrace{777...7}}$$$$\mathrm{has}\mathrm{remainder}\mathrm{of}7\mathrm{when}\mathrm{divided}\mathrm{by}9$$$$$$

Answered by floor(10²Eta[1]) last updated on 23/Dec/20

$$\equiv 7+7.2+7.3+...+7.2020\left(\mathrm{mod}9\right)$$$$=7(1+2+3+...+2020)=7.2021.1010$$$$=7.(2016+5)(1008+2)\equiv 7.5.2=70\equiv 7\left(\mathrm{mod}9\right)$$

Commented by talminator2856791 last updated on 24/Dec/20

$$\mathrm{how}\mathrm{can}\mathrm{you}\mathrm{say}$$$$\equiv 7+7.2+7.3+...+7.2020\left(\mathrm{mod}9\right)?$$$$$$

Commented by floor(10²Eta[1]) last updated on 24/Dec/20

$${\mathrm{a}}_{\mathrm{n}}{10}^{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}{10}^{\mathrm{n}-1}+...+{\mathrm{a}}_{1}10+{\mathrm{a}}_0$$$$\equiv {\mathrm{a}}_{\mathrm{n}}+{\mathrm{a}}_{\mathrm{n}-1}+...+{\mathrm{a}}_1+{\mathrm{a}}_0\left(\mathrm{mod}9\right)$$$$\mathrm{so}77\equiv 7+7=7.2\left(\mathrm{mod}9\right)\mathrm{because}77=7.10+7$$$$\mathrm{and}\mathrm{so}\mathrm{on}...$$

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