if-11-800-divide-by-9-what-will-be-the-remainder-

Question Number 203911 by Davidtim last updated on 01/Feb/24

$${if}\:\mathrm{11}^{\mathrm{800}} \:{divide}\:{by}\:\mathrm{9}\:{what}\:{will}\:{be}\:{the}\: \\ $$$${remainder}? \\ $$

Answered by AST last updated on 01/Feb/24

11^(800) ≡2^(800) =(2^6 )^(133) (2)^2 ≡4(mod 9)

$$\mathrm{11}^{\mathrm{800}} \equiv\mathrm{2}^{\mathrm{800}} =\left(\mathrm{2}^{\mathrm{6}} \right)^{\mathrm{133}} \left(\mathrm{2}\right)^{\mathrm{2}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right) \\ $$

Answered by Rasheed.Sindhi last updated on 02/Feb/24

In some detail: 11^(800) ≡2^(800) ≡x(mod 9) (say) ∵ gcd(2,9)=1 ∴ 2^(φ(9)) ≡1(mod 9) 2^6 ≡1(mod 9) (2^6 )^(133) ≡1^(133) (mod 9) 2^(798) ≡1(mod 9) 2^(798) .2^2 ≡1.2^2 (mod 9) 2^(800) ≡4(mod 9) x≡4(mod 9)

$${In}\:{some}\:{detail}: \\ $$$$\mathrm{11}^{\mathrm{800}} \equiv\mathrm{2}^{\mathrm{800}} \equiv{x}\left({mod}\:\mathrm{9}\right)\:\left({say}\right) \\ $$$$\because\:\mathrm{gcd}\left(\mathrm{2},\mathrm{9}\right)=\mathrm{1} \\ $$$$\therefore\:\:\:\:\:\mathrm{2}^{\phi\left(\mathrm{9}\right)} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\left(\mathrm{2}^{\mathrm{6}} \right)^{\mathrm{133}} \equiv\mathrm{1}^{\mathrm{133}} \left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{798}} \equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{798}} .\mathrm{2}^{\mathrm{2}} \equiv\mathrm{1}.\mathrm{2}^{\mathrm{2}} \left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{800}} \equiv\mathrm{4}\left({mod}\:\mathrm{9}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{x}\equiv\mathrm{4}\left({mod}\:\mathrm{9}\right) \\ $$

Commented by Davidtim last updated on 05/Feb/24

$${I}\:{didn}'{t}\:{get}\:{full}\:{result},\:{would}\:\:{you}\:{mind}\:{solve}\:{desribely}? \\ $$

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