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Question-9325




Question Number 9325 by tawakalitu last updated on 30/Nov/16
Answered by RasheedSoomro last updated on 30/Nov/16
(B) 12   25≡−1(mod 13)  (25)^(15) ≡(−1)^(15) (mod 13)  (25)^(15) ≡−1(mod 13)  (25)^(15) ≡−1+13(mod 13)  (25)^(15) ≡12(mod 13)
$$\left(\mathrm{B}\right)\:\mathrm{12}\: \\ $$$$\mathrm{25}\equiv−\mathrm{1}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}} \equiv\left(−\mathrm{1}\right)^{\mathrm{15}} \left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}} \equiv−\mathrm{1}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}} \equiv−\mathrm{1}+\mathrm{13}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$$$\left(\mathrm{25}\right)^{\mathrm{15}} \equiv\mathrm{12}\left(\mathrm{mod}\:\mathrm{13}\right) \\ $$
Commented by tawakalitu last updated on 30/Nov/16
Thanks so much sir.
$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}. \\ $$

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