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2009-3-2016n-2013-2010-2-2016n-2013-x-mod-11-where-n-is-any-integer-0-

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Question Number 192275 by York12 last updated on 13/May/23
2009^3^(2016n+2013) +2010^2^(2016n+2013) ≡x mod(11) where n is any integer ≥0
$$\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } +\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \equiv{x}\:{mod}\left(\mathrm{11}\right)\:{where}\:{n}\:{is}\:{any}\:{integer}\:\geq\mathrm{0} \\ $$$$ \\ $$
Answered by BaliramKumar last updated on 14/May/23
0
$$\mathrm{0} \\ $$
Answered by BaliramKumar last updated on 14/May/23
φ(11) = 10 ⇒ φ(10) = 4 ((2009)/(11))=7 rem.⇒(3/(10))= 3 rem. ⇒((2016n+2013)/4) = 1 rem. ((2010)/(11))=8 rem.⇒(2/(10))= 2 rem. ⇒((2016n+2013)/4) = 1 rem. (7^3^1 /(11)) + (8^2^1 /(11)) = ((343)/(11)) + ((64)/(11)) = (2/(11)) + ((−2)/(11)) = (0/(11)) = 0 Remainder
$$\phi\left(\mathrm{11}\right)\:=\:\mathrm{10}\:\Rightarrow\:\phi\left(\mathrm{10}\right)\:=\:\mathrm{4} \\ $$$$\frac{\mathrm{2009}}{\mathrm{11}}=\mathrm{7}\:\mathrm{rem}.\Rightarrow\frac{\mathrm{3}}{\mathrm{10}}=\:\mathrm{3}\:\mathrm{rem}.\:\Rightarrow\frac{\mathrm{2016n}+\mathrm{2013}}{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{rem}.\:\:\:\:\:\:\: \\ $$$$\frac{\mathrm{2010}}{\mathrm{11}}=\mathrm{8}\:\mathrm{rem}.\Rightarrow\frac{\mathrm{2}}{\mathrm{10}}=\:\mathrm{2}\:\mathrm{rem}.\:\Rightarrow\frac{\mathrm{2016n}+\mathrm{2013}}{\mathrm{4}}\:=\:\mathrm{1}\:\mathrm{rem}.\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{\mathrm{7}^{\mathrm{3}^{\mathrm{1}} } }{\mathrm{11}}\:\:+\:\frac{\mathrm{8}^{\mathrm{2}^{\mathrm{1}} } }{\mathrm{11}}\:=\:\frac{\mathrm{343}}{\mathrm{11}}\:+\:\frac{\mathrm{64}}{\mathrm{11}}\:=\:\frac{\mathrm{2}}{\mathrm{11}}\:+\:\frac{−\mathrm{2}}{\mathrm{11}}\:=\:\frac{\mathrm{0}}{\mathrm{11}}\:=\:\mathrm{0}\:\mathrm{Remainder} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by York12 last updated on 14/May/23
2009 ≡ 7 mod (11) → 2009^(10) ≡ 7^(10) mod (11) ∵ 11 is a prime number → we can apply fermat′s little theorm ∴ 7^(10) ≡ 1 mod (11) → 7^(10k) ≡ 1 mod (11) where k is a an integer ≥ 0 2009^3^(2016n+2013) ≡ 7^3^(2016n+2013) mod (11) 3^(2016n+2013) can be written as 3^(4(504n+503)+1) → The Unit digit would be 3 ∴ 3^(2016n+2013) ≡ 3 mod (10) → 3^(2016n+2013) can be written as 10k +3 7^(10k) ≡ 1 mod (11) , 7^3 ≡ 2 mod (11) ∴ 7^(10k+3) ≡ 2 mod (11) ∴ 2009^3^(2016n+2013) ≡ 2 mod (11) → {I} 2010 ≡ 8 mod (11) → 2010^(10) ≡ 8^(10) mod (11) By using fermat′s little theorm: we can state the following 8^(10 ) ≡ 1 mod (11) → 8^(10m) ≡ 1 mod (11) 2010^2^(2016n+2013) ≡ 8^2^(2016n+2013) mod(11) 2^(2016n+2013) can be written as 2^(4(504n+503)+1) ∴The Unit digit of 2^(2016n+2013) is 2 ∴2^(2016n+2013) ≡ 2 mod (10) → can bd written as 10m+2 8^(10m) ≡ 1 mod (11) , 8^2 ≡ 9 mod (11) ∴ 8^(10m+2) ≡ 9 mod (11) ∴ 2010^2^(2016n+2013) ≡ 9 mod (11) → {II} from I and II we conclude that: 2009^3^(2016n+2013) + 2010^2^(2016n+2013) ≡ 0 mod (11) ∴ x = 0 (That′s it) [BY YORK]
$$ \\ $$$$\mathrm{2009}\:\equiv\:\mathrm{7}\:{mod}\:\:\left(\mathrm{11}\right)\:\:\:\rightarrow\:\:\mathrm{2009}^{\mathrm{10}} \:\equiv\:\mathrm{7}^{\mathrm{10}} \:{mod}\:\left(\mathrm{11}\right)\: \\ $$$$\because\:\mathrm{11}\:{is}\:{a}\:{prime}\:{number}\:\rightarrow\:{we}\:{can}\:{apply}\:{fermat}'{s}\:{little}\:\:{theorm} \\ $$$$\therefore\:\mathrm{7}^{\mathrm{10}} \:\equiv\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right)\:\:\:\rightarrow\:\mathrm{7}^{\mathrm{10}{k}} \:\equiv\:\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right)\: \\ $$$${where}\:{k}\:{is}\:{a}\:\:{an}\:\:{integer}\:\geq\:\mathrm{0} \\ $$$$\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } \:\equiv\:\mathrm{7}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } \:{mod}\:\left(\mathrm{11}\right) \\ $$$$\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} \:\:{can}\:{be}\:{written}\:{as}\:\mathrm{3}^{\mathrm{4}\left(\mathrm{504}{n}+\mathrm{503}\right)+\mathrm{1}} \:\rightarrow\:{The}\:{Unit}\:{digit}\:{would}\:{be}\:\mathrm{3} \\ $$$$\therefore\:\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} \:\equiv\:\mathrm{3}\:{mod}\:\left(\mathrm{10}\right)\:\rightarrow\:\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} \:\:{can}\:{be}\:{written}\:{as}\:\mathrm{10}{k}\:\:+\mathrm{3} \\ $$$$ \\ $$$$\mathrm{7}^{\mathrm{10}{k}} \:\equiv\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right)\:,\:\:\mathrm{7}^{\mathrm{3}} \:\equiv\:\mathrm{2}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\therefore\:\:\mathrm{7}^{\mathrm{10}{k}+\mathrm{3}} \:\equiv\:\mathrm{2}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\therefore\:\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}\boldsymbol{{n}}+\mathrm{2013}} } \:\equiv\:\mathrm{2}\:{mod}\:\left(\mathrm{11}\right)\:\rightarrow\:\:\left\{\boldsymbol{\mathcal{I}}\right\} \\ $$$$\mathrm{2010}\:\equiv\:\mathrm{8}\:{mod}\:\left(\mathrm{11}\right)\:\rightarrow\:\mathrm{2010}^{\mathrm{10}} \:\equiv\:\mathrm{8}^{\mathrm{10}} \:{mod}\:\left(\mathrm{11}\right) \\ $$$${By}\:{using}\:{fermat}'{s}\:{little}\:{theorm}:\:{we}\:{can}\:{state}\:{the}\:{following}\: \\ $$$$\mathrm{8}^{\mathrm{10}\:} \:\equiv\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right)\:\rightarrow\:\mathrm{8}^{\mathrm{10}{m}} \:\equiv\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \:\equiv\:\:\mathrm{8}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \:{mod}\left(\mathrm{11}\right) \\ $$$$\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} \:{can}\:{be}\:{written}\:{as}\:\mathrm{2}^{\mathrm{4}\left(\mathrm{504}{n}+\mathrm{503}\right)+\mathrm{1}} \\ $$$$\therefore{The}\:{Unit}\:{digit}\:{of}\:\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} \:{is}\:\mathrm{2} \\ $$$$\therefore\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} \:\equiv\:\mathrm{2}\:{mod}\:\left(\mathrm{10}\right)\:\rightarrow\:{can}\:{bd}\:{written}\:{as}\:\mathrm{10}{m}+\mathrm{2}\: \\ $$$$\mathrm{8}^{\mathrm{10}{m}} \:\equiv\:\mathrm{1}\:{mod}\:\left(\mathrm{11}\right)\:,\:\mathrm{8}^{\mathrm{2}} \:\equiv\:\mathrm{9}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\therefore\:\mathrm{8}^{\mathrm{10}{m}+\mathrm{2}} \:\equiv\:\:\mathrm{9}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\therefore\:\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \:\equiv\:\mathrm{9}\:{mod}\:\left(\mathrm{11}\right)\:\rightarrow\:\left\{\mathcal{II}\right\} \\ $$$${from}\:\mathcal{I}\:{and}\:\mathcal{II}\:{we}\:{conclude}\:{that}: \\ $$$$ \\ $$$$\mathrm{2009}^{\mathrm{3}^{\mathrm{2016}{n}+\mathrm{2013}} } \:+\:\mathrm{2010}^{\mathrm{2}^{\mathrm{2016}{n}+\mathrm{2013}} } \:\equiv\:\mathrm{0}\:{mod}\:\left(\mathrm{11}\right) \\ $$$$\therefore\:\boldsymbol{{x}}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{{T}}{hat}'{s}\:{it}\right) \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\mathscr{BY}\:\:{YORK}\right] \\ $$

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