Question-57075

Question Number 57075 by Tawa1 last updated on 30/Mar/19

Commented by Tawa1 last updated on 30/Mar/19

Please i want to understand how the −1 in the expansion becomes − 13 and the + 60 outside. and how the remainder is 8. Using the same method.

$$\mathrm{Please}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{the}\:−\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{becomes}\:−\:\mathrm{13} \\ $$$$\mathrm{and}\:\mathrm{the}\:+\:\mathrm{60}\:\mathrm{outside}.\:\mathrm{and}\:\mathrm{how}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\:\mathrm{8}.\:\:\mathrm{Using}\:\mathrm{the}\:\mathrm{same}\:\mathrm{method}. \\ $$

Commented by 121194 last updated on 30/Mar/19

−1=12−13 5×12=60 since 13∣26 then 5×(−13)+52+8 13k+52+8 and 13∣52 so you can easy deduct that 5^(99) %13=8

$$−\mathrm{1}=\mathrm{12}−\mathrm{13} \\ $$$$\mathrm{5}×\mathrm{12}=\mathrm{60} \\ $$$$\mathrm{since} \\ $$$$\mathrm{13}\mid\mathrm{26} \\ $$$$\mathrm{then} \\ $$$$\mathrm{5}×\left(−\mathrm{13}\right)+\mathrm{52}+\mathrm{8} \\ $$$$\mathrm{13}{k}+\mathrm{52}+\mathrm{8} \\ $$$$\mathrm{and} \\ $$$$\mathrm{13}\mid\mathrm{52} \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{can}\:\mathrm{easy}\:\mathrm{deduct}\:\mathrm{that}\:\mathrm{5}^{\mathrm{99}} \%\mathrm{13}=\mathrm{8} \\ $$

Commented by Tawa1 last updated on 30/Mar/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

Commented by JDamian last updated on 30/Mar/19

5^(99) =5(26−1)^(49) (5^(99) )mod 13=[5(26−1)^(49) ]mod 13= =(5 mod 13)×(26−1)^(49) mod 13= =5×(26 mod 13−1 mod 13)^(49) = =5×(0−1)^(49) =−5 −5 mod 13=(−5+13) mod 13=8 mod 13= =8

$$\mathrm{5}^{\mathrm{99}} =\mathrm{5}\left(\mathrm{26}−\mathrm{1}\right)^{\mathrm{49}} \\ $$$$ \\ $$$$\left(\mathrm{5}^{\mathrm{99}} \right){mod}\:\mathrm{13}=\left[\mathrm{5}\left(\mathrm{26}−\mathrm{1}\right)^{\mathrm{49}} \right]{mod}\:\mathrm{13}= \\ $$$$=\left(\mathrm{5}\:{mod}\:\mathrm{13}\right)×\left(\mathrm{26}−\mathrm{1}\right)^{\mathrm{49}} {mod}\:\mathrm{13}= \\ $$$$=\mathrm{5}×\left(\mathrm{26}\:{mod}\:\mathrm{13}−\mathrm{1}\:{mod}\:\mathrm{13}\right)^{\mathrm{49}} = \\ $$$$=\mathrm{5}×\left(\mathrm{0}−\mathrm{1}\right)^{\mathrm{49}} =−\mathrm{5} \\ $$$$−\mathrm{5}\:{mod}\:\mathrm{13}=\left(−\mathrm{5}+\mathrm{13}\right)\:{mod}\:\mathrm{13}=\mathrm{8}\:{mod}\:\mathrm{13}= \\ $$$$=\mathrm{8} \\ $$

Commented by Tawa1 last updated on 30/Mar/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19

5^(99) (5^3 )^(33) (125)^(33) (117+8)^(33) (9×13+8)^(33) (9^ ×13)^(33) +33c_1 (9×13)^(32) (8)+...+8^(33) f(13)+(8^3 )^(11) f(13)+(507+5)^(11) f(13)+(39×13+5)^(11) f(13)+{(39×13)^(11) +11c_1 (39×13)^(10) ×5+...+5^(11) } f(13)+g(13)+5^(11) f(13)+g(13)+(5^3 )^3 (5^2 ) f(13)+g(13)+(117+8)^3 (25) f(13)+g(13)+25[(13×9)^3 +3(13×9)^2 ×8+3(13×9)×8^2 +8^3 ] f(13)+g(13)+25h(13)+25×8^3 now calculating 25×8^3 (13+12)(507+5) (13+12)(39×13+5) 13×39×13+13×5+12×39×13+5×12 ϕ(13)+(4×13+8) so remainder is 8

$$\mathrm{5}^{\mathrm{99}} \\ $$$$\left(\mathrm{5}^{\mathrm{3}} \right)^{\mathrm{33}} \\ $$$$\left(\mathrm{125}\right)^{\mathrm{33}} \\ $$$$\left(\mathrm{117}+\mathrm{8}\right)^{\mathrm{33}} \\ $$$$\left(\mathrm{9}×\mathrm{13}+\mathrm{8}\right)^{\mathrm{33}} \\ $$$$\left(\mathrm{9}^{} ×\mathrm{13}\right)^{\mathrm{33}} +\mathrm{33}{c}_{\mathrm{1}} \left(\mathrm{9}×\mathrm{13}\right)^{\mathrm{32}} \left(\mathrm{8}\right)+…+\mathrm{8}^{\mathrm{33}} \\ $$$${f}\left(\mathrm{13}\right)+\left(\mathrm{8}^{\mathrm{3}} \right)^{\mathrm{11}} \\ $$$${f}\left(\mathrm{13}\right)+\left(\mathrm{507}+\mathrm{5}\right)^{\mathrm{11}} \\ $$$${f}\left(\mathrm{13}\right)+\left(\mathrm{39}×\mathrm{13}+\mathrm{5}\right)^{\mathrm{11}} \\ $$$${f}\left(\mathrm{13}\right)+\left\{\left(\mathrm{39}×\mathrm{13}\right)^{\mathrm{11}} +\mathrm{11}{c}_{\mathrm{1}} \left(\mathrm{39}×\mathrm{13}\right)^{\mathrm{10}} ×\mathrm{5}+…+\mathrm{5}^{\mathrm{11}} \right\} \\ $$$${f}\left(\mathrm{13}\right)+{g}\left(\mathrm{13}\right)+\mathrm{5}^{\mathrm{11}} \\ $$$${f}\left(\mathrm{13}\right)+{g}\left(\mathrm{13}\right)+\left(\mathrm{5}^{\mathrm{3}} \right)^{\mathrm{3}} \left(\mathrm{5}^{\mathrm{2}} \right) \\ $$$${f}\left(\mathrm{13}\right)+{g}\left(\mathrm{13}\right)+\left(\mathrm{117}+\mathrm{8}\right)^{\mathrm{3}} \left(\mathrm{25}\right) \\ $$$${f}\left(\mathrm{13}\right)+{g}\left(\mathrm{13}\right)+\mathrm{25}\left[\left(\mathrm{13}×\mathrm{9}\right)^{\mathrm{3}} +\mathrm{3}\left(\mathrm{13}×\mathrm{9}\right)^{\mathrm{2}} ×\mathrm{8}+\mathrm{3}\left(\mathrm{13}×\mathrm{9}\right)×\mathrm{8}^{\mathrm{2}} +\mathrm{8}^{\mathrm{3}} \right] \\ $$$${f}\left(\mathrm{13}\right)+{g}\left(\mathrm{13}\right)+\mathrm{25}{h}\left(\mathrm{13}\right)+\mathrm{25}×\mathrm{8}^{\mathrm{3}} \\ $$$${now}\:{calculating}\:\mathrm{25}×\mathrm{8}^{\mathrm{3}} \: \\ $$$$\left(\mathrm{13}+\mathrm{12}\right)\left(\mathrm{507}+\mathrm{5}\right) \\ $$$$\left(\mathrm{13}+\mathrm{12}\right)\left(\mathrm{39}×\mathrm{13}+\mathrm{5}\right) \\ $$$$\mathrm{13}×\mathrm{39}×\mathrm{13}+\mathrm{13}×\mathrm{5}+\mathrm{12}×\mathrm{39}×\mathrm{13}+\mathrm{5}×\mathrm{12} \\ $$$$\varphi\left(\mathrm{13}\right)+\left(\mathrm{4}×\mathrm{13}+\mathrm{8}\right) \\ $$$${so}\:{remainder}\:{is}\:\mathrm{8} \\ $$$$ \\ $$

Commented by peter frank last updated on 30/Mar/19