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Question Number 6843 by Tawakalitu. last updated on 30/Jul/16

2^(10) + 3^(20) + 4^(30) + 5^(40) + 6^(50) (mod 7) = ?

$$\mathrm{2}^{\mathrm{10}} \:+\:\mathrm{3}^{\mathrm{20}} \:+\:\mathrm{4}^{\mathrm{30}} \:+\:\mathrm{5}^{\mathrm{40}} \:+\:\mathrm{6}^{\mathrm{50}} \:\:\left({mod}\:\mathrm{7}\right)\:=\:? \\ $$

Commented by prakash jain last updated on 31/Jul/16

2^3 ≡ 1 (mod 7)⇒2^(10) =2^3 ×2^3 ×2^3 ×2≡2 (mod 7) 3^2 =9≡2 (mod 7)⇒3^(20) ≡2^(10) ≡2 (mod 7) 4^(30) =(2^(10) )^6 ≡2^6 ≡1 (mod 7) 5^2 ≡4 (mod 7)⇒5^(40) ≡4^(20) ≡(2^(10) )^4 ≡2^4 ≡2 (mod 7) 6≡−1 (mod 7)⇒6^(50) ≡(−1)^(50) ≡1 (mod 7) 2^(10) + 3^(20) + 4^(30) + 5^(40) + 6^(50) (mod 7) = 2+2+1+2+1≡8≡1 (mod 7)

$$\mathrm{2}^{\mathrm{3}} \:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{2}^{\mathrm{10}} =\mathrm{2}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} ×\mathrm{2}\equiv\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{3}^{\mathrm{2}} =\mathrm{9}\equiv\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{3}^{\mathrm{20}} \equiv\mathrm{2}^{\mathrm{10}} \equiv\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{4}^{\mathrm{30}} =\left(\mathrm{2}^{\mathrm{10}} \right)^{\mathrm{6}} \equiv\mathrm{2}^{\mathrm{6}} \equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{5}^{\mathrm{2}} \equiv\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{5}^{\mathrm{40}} \equiv\mathrm{4}^{\mathrm{20}} \equiv\left(\mathrm{2}^{\mathrm{10}} \right)^{\mathrm{4}} \equiv\mathrm{2}^{\mathrm{4}} \equiv\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{6}\equiv−\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right)\Rightarrow\mathrm{6}^{\mathrm{50}} \equiv\left(−\mathrm{1}\right)^{\mathrm{50}} \equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$$$\mathrm{2}^{\mathrm{10}} \:+\:\mathrm{3}^{\mathrm{20}} \:+\:\mathrm{4}^{\mathrm{30}} \:+\:\mathrm{5}^{\mathrm{40}} \:+\:\mathrm{6}^{\mathrm{50}} \:\:\left({mod}\:\mathrm{7}\right)\:=\: \\ $$$$\mathrm{2}+\mathrm{2}+\mathrm{1}+\mathrm{2}+\mathrm{1}\equiv\mathrm{8}\equiv\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$

Commented by Tawakalitu. last updated on 31/Jul/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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