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

Find the remainder 7+77+777+7777+...+777...7_(2020 times) when divided by 9.

$$\:{Find}\:{the}\:{remainder}\: \\ $$$$\:\mathrm{7}+\mathrm{77}+\mathrm{777}+\mathrm{7777}+...+\underset{\mathrm{2020}\:{times}} {\underbrace{\mathrm{777}...\mathrm{7}}} \\ $$$${when}\:{divided}\:{by}\:\mathrm{9}. \\ $$

Answered by talminator2856791 last updated on 23/Dec/20

7Σ_(k=1) ^(2020) k(10^(2020−k) ) = 7Σ_(k=1) ^(2020) Σ_(i=1) ^k 10^(i−1) 10^n ≡ 1 mod 9 , n ∈ N ⇒ ((7Σ_(k=1) ^(2020) Σ_(i=1) ^k 10^(i−1) )/9) = 9j + 7(2020)+ 7Σ_(k=1) ^(2019) k = 9j + 14140 + 7(2019)(1010) = 9j + 14 288 470 ((14 288 470)/9) = 1 587 607.7777..... ⇒ 7+77+777+.......+777...7_(2020 times) has remainder of 7 when divided by 9

$$\: \\ $$$$\:\mathrm{7}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\sum}}\:{k}\left(\mathrm{10}^{\mathrm{2020}−{k}} \right)\:=\:\mathrm{7}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\:\mathrm{10}^{{i}−\mathrm{1}} \\ $$$$\: \\ $$$$\:\mathrm{10}^{{n}} \:\equiv\:\mathrm{1}\:\mathrm{mod}\:\mathrm{9}\:,\:\:{n}\:\in\:\mathbb{N} \\ $$$$\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{7}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2020}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\:\mathrm{10}^{{i}−\mathrm{1}} }{\mathrm{9}}\:=\:\mathrm{9}{j}\:+\:\mathrm{7}\left(\mathrm{2020}\right)+\:\mathrm{7}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2019}} {\sum}}{k}\: \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{9}{j}\:+\:\mathrm{14140}\:+\:\mathrm{7}\left(\mathrm{2019}\right)\left(\mathrm{1010}\right) \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{9}{j}\:+\:\mathrm{14}\:\mathrm{288}\:\mathrm{470}\: \\ $$$$\: \\ $$$$\:\frac{\mathrm{14}\:\mathrm{288}\:\mathrm{470}}{\mathrm{9}}\:=\:\mathrm{1}\:\mathrm{587}\:\mathrm{607}.\mathrm{7777}..... \\ $$$$\: \\ $$$$\:\Rightarrow\:\mathrm{7}+\mathrm{77}+\mathrm{777}+.......+\underset{\mathrm{2020}\:\mathrm{times}} {\underbrace{\mathrm{777}...\mathrm{7}}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{has}\:\mathrm{remainder}\:\mathrm{of}\:\mathrm{7}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{9} \\ $$$$\: \\ $$

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

≡7+7.2+7.3+...+7.2020(mod9) =7(1+2+3+...+2020)=7.2021.1010 =7.(2016+5)(1008+2)≡7.5.2=70≡7(mod9)

$$\equiv\mathrm{7}+\mathrm{7}.\mathrm{2}+\mathrm{7}.\mathrm{3}+...+\mathrm{7}.\mathrm{2020}\left(\mathrm{mod9}\right) \\ $$$$=\mathrm{7}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{2020}\right)=\mathrm{7}.\mathrm{2021}.\mathrm{1010} \\ $$$$=\mathrm{7}.\left(\mathrm{2016}+\mathrm{5}\right)\left(\mathrm{1008}+\mathrm{2}\right)\equiv\mathrm{7}.\mathrm{5}.\mathrm{2}=\mathrm{70}\equiv\mathrm{7}\left(\mathrm{mod9}\right) \\ $$

Commented by talminator2856791 last updated on 24/Dec/20

how can you say ≡7+7.2+7.3+...+7.2020(mod9)?

$$\:\mathrm{how}\:\mathrm{can}\:\mathrm{you}\:\mathrm{say} \\ $$$$\:\equiv\mathrm{7}+\mathrm{7}.\mathrm{2}+\mathrm{7}.\mathrm{3}+...+\mathrm{7}.\mathrm{2020}\left(\mathrm{mod9}\right)? \\ $$$$\: \\ $$

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

a_n 10^n +a_(n−1) 10^(n−1) +...+a_1 10+a_0 ≡a_n +a_(n−1) +...+a_1 +a_0 (mod 9) so 77≡7+7=7.2(mod 9) because 77=7.10+7 and so on...

$$\mathrm{a}_{\mathrm{n}} \mathrm{10}^{\mathrm{n}} +\mathrm{a}_{\mathrm{n}−\mathrm{1}} \mathrm{10}^{\mathrm{n}−\mathrm{1}} +...+\mathrm{a}_{\mathrm{1}} \mathrm{10}+\mathrm{a}_{\mathrm{0}} \\ $$$$\equiv\mathrm{a}_{\mathrm{n}} +\mathrm{a}_{\mathrm{n}−\mathrm{1}} +...+\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{0}} \left(\mathrm{mod}\:\mathrm{9}\right) \\ $$$$\mathrm{so}\:\mathrm{77}\equiv\mathrm{7}+\mathrm{7}=\mathrm{7}.\mathrm{2}\left(\mathrm{mod}\:\mathrm{9}\right)\:\mathrm{because}\:\mathrm{77}=\mathrm{7}.\mathrm{10}+\mathrm{7} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}... \\ $$

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