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6-2020-8-2020-mod-49-Show-your-elegant-workings-please-Thanks-a-lot-

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Question Number 121608 by naka3546 last updated on 10/Nov/20
(6^(2020) + 8^(2020) ) mod 49 ? Show your elegant workings , please. Thanks a lot.
$$\left(\mathrm{6}^{\mathrm{2020}} \:+\:\mathrm{8}^{\mathrm{2020}} \right)\:{mod}\:\mathrm{49}\:\:? \\ $$$${Show}\:\:{your}\:\:{elegant}\:\:{workings}\:,\:{please}. \\ $$$${Thanks}\:\:{a}\:{lot}. \\ $$
Answered by mr W last updated on 10/Nov/20
6^(2020) +8^(2020) =(7−1)^(2020) +(7+1)^(2020) =(1−7)^(2020) +(1+7)^(2020) =Σ_(k=0) ^(2020) C_k ^(2020) (−1)^k 7^k +Σ_(k=0) ^(2020) C_k ^(2020) 7^k =Σ_(k=0,2,...) ^(2020) 2C_k ^(2020) 7^k =2+2Σ_(k=1) ^(1010) C_(2k) ^(2020) 7^(2k) =2+2Σ_(k=1) ^(1010) C_(2k) ^(2020) 49^k ⇒(6^(2020) +8^(2020) ) mod 49=2
$$\mathrm{6}^{\mathrm{2020}} +\mathrm{8}^{\mathrm{2020}} \\ $$$$=\left(\mathrm{7}−\mathrm{1}\right)^{\mathrm{2020}} +\left(\mathrm{7}+\mathrm{1}\right)^{\mathrm{2020}} \\ $$$$=\left(\mathrm{1}−\mathrm{7}\right)^{\mathrm{2020}} +\left(\mathrm{1}+\mathrm{7}\right)^{\mathrm{2020}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}{C}_{{k}} ^{\mathrm{2020}} \left(−\mathrm{1}\right)^{{k}} \mathrm{7}^{{k}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{2020}} {\sum}}{C}_{{k}} ^{\mathrm{2020}} \mathrm{7}^{{k}} \\ $$$$=\underset{{k}=\mathrm{0},\mathrm{2},…} {\overset{\mathrm{2020}} {\sum}}\mathrm{2}{C}_{{k}} ^{\mathrm{2020}} \mathrm{7}^{{k}} \\ $$$$=\mathrm{2}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1010}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2020}} \mathrm{7}^{\mathrm{2}{k}} \\ $$$$=\mathrm{2}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{1010}} {\sum}}{C}_{\mathrm{2}{k}} ^{\mathrm{2020}} \mathrm{49}^{{k}} \\ $$$$\Rightarrow\left(\mathrm{6}^{\mathrm{2020}} +\mathrm{8}^{\mathrm{2020}} \right)\:{mod}\:\mathrm{49}=\mathrm{2} \\ $$
Answered by Rasheed.Sindhi last updated on 10/Nov/20
≡ An Other Way ≡ ^• 6^7 ≡−1(mod49) 2020=7×288+4 (6^7 )^(288) (6)^4 ≡(−1)^(144) (−1)^4 (mod49) 6^(2020) ≡1(mod49)....................A ^• 8^7 ≡1(mod49) (8^7 )^(288) (8)^4 ≡(1)^(288) (1)^4 (mod49) 8^(2020) ≡1(mod 49).................B A+B: 6^(2020) +8^(2020) ≡2(mod49)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv\:\mathbb{A}\mathrm{n}\:\mathbb{O}\mathrm{ther}\:\mathbb{W}\mathrm{ay}\:\equiv \\ $$$$\:^{\bullet} \mathrm{6}^{\mathrm{7}} \equiv−\mathrm{1}\left({mod}\mathrm{49}\right)\: \\ $$$$\:\:\mathrm{2020}=\mathrm{7}×\mathrm{288}+\mathrm{4} \\ $$$$\left(\mathrm{6}^{\mathrm{7}} \right)^{\mathrm{288}} \left(\mathrm{6}\right)^{\mathrm{4}} \equiv\left(−\mathrm{1}\right)^{\mathrm{144}} \left(−\mathrm{1}\right)^{\mathrm{4}} \left({mod}\mathrm{49}\right) \\ $$$$\mathrm{6}^{\mathrm{2020}} \equiv\mathrm{1}\left({mod}\mathrm{49}\right)………………..{A} \\ $$$$ \\ $$$$\:^{\bullet} \mathrm{8}^{\mathrm{7}} \equiv\mathrm{1}\left({mod}\mathrm{49}\right) \\ $$$$\left(\mathrm{8}^{\mathrm{7}} \right)^{\mathrm{288}} \left(\mathrm{8}\right)^{\mathrm{4}} \equiv\left(\mathrm{1}\right)^{\mathrm{288}} \left(\mathrm{1}\right)^{\mathrm{4}} \left({mod}\mathrm{49}\right) \\ $$$$\mathrm{8}^{\mathrm{2020}} \equiv\mathrm{1}\left({mod}\:\mathrm{49}\right)……………..{B} \\ $$$${A}+{B}: \\ $$$$\mathrm{6}^{\mathrm{2020}} +\mathrm{8}^{\mathrm{2020}} \equiv\mathrm{2}\left({mod}\mathrm{49}\right) \\ $$

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