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The-remainder-of-13-13-when-divided-by-99-is-




Question Number 169159 by SLVR last updated on 25/Apr/22
The remainder of  13^(13) when   divided by 99 is
$${The}\:{remainder}\:{of} \\ $$$$\mathrm{13}^{\mathrm{13}} {when}\: \\ $$$${divided}\:{by}\:\mathrm{99}\:{is} \\ $$
Commented by SLVR last updated on 25/Apr/22
kindly solve..
$${kindly}\:{solve}.. \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 25/Apr/22
13^2 ≡169(mod 99)  13^2 ≡169−99=70(mod 99)  13^2 ×13≡70×13(mod 99)  13^3 ≡910(mod 99)  13^3 ≡910−99×9=19 (mod 99)  (13^3 )^2 ≡19^2 =361(mod 99)  13^6 ≡361−99×3=64(mod 99)  (13^6 )^2 ≡64^2 =4096(mod 99)  13^(12) ≡4096−99×41=37(mod 99)  13^(12) ∙13≡37×13=481(mod 99)  13^(13) ≡481−99×4=85(mod 99)  1y3^(13) ≡85(mod 99)
$$\mathrm{13}^{\mathrm{2}} \equiv\mathrm{169}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{2}} \equiv\mathrm{169}−\mathrm{99}=\mathrm{70}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{2}} ×\mathrm{13}\equiv\mathrm{70}×\mathrm{13}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{3}} \equiv\mathrm{910}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{3}} \equiv\mathrm{910}−\mathrm{99}×\mathrm{9}=\mathrm{19}\:\left({mod}\:\mathrm{99}\right) \\ $$$$\left(\mathrm{13}^{\mathrm{3}} \right)^{\mathrm{2}} \equiv\mathrm{19}^{\mathrm{2}} =\mathrm{361}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{6}} \equiv\mathrm{361}−\mathrm{99}×\mathrm{3}=\mathrm{64}\left({mod}\:\mathrm{99}\right) \\ $$$$\left(\mathrm{13}^{\mathrm{6}} \right)^{\mathrm{2}} \equiv\mathrm{64}^{\mathrm{2}} =\mathrm{4096}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{12}} \equiv\mathrm{4096}−\mathrm{99}×\mathrm{41}=\mathrm{37}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{12}} \centerdot\mathrm{13}\equiv\mathrm{37}×\mathrm{13}=\mathrm{481}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{13}^{\mathrm{13}} \equiv\mathrm{481}−\mathrm{99}×\mathrm{4}=\mathrm{85}\left({mod}\:\mathrm{99}\right) \\ $$$$\mathrm{1}{y}\mathrm{3}^{\mathrm{13}} \equiv\mathrm{85}\left({mod}\:\mathrm{99}\right) \\ $$
Commented by SLVR last updated on 25/Apr/22
So..kind of you sir...
$${So}..{kind}\:{of}\:{you}\:{sir}… \\ $$

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