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Question-211453




Question Number 211453 by BaliramKumar last updated on 09/Sep/24
Answered by A5T last updated on 09/Sep/24
5^a ≡5(mod 10);11^c ≡1(mod 10)  ⇒5^a +11^c ≡6(mod 10)  7^b ≡7,9,3 or 1 (mod 10);13^d ≡3,9,7 or 1(mod 10)  ⇒5^a +7^b +11^c +13^d ≡6;2;8;4;0 (mod 10)⇒(c)
$$\mathrm{5}^{{a}} \equiv\mathrm{5}\left({mod}\:\mathrm{10}\right);\mathrm{11}^{{c}} \equiv\mathrm{1}\left({mod}\:\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{5}^{{a}} +\mathrm{11}^{{c}} \equiv\mathrm{6}\left({mod}\:\mathrm{10}\right) \\ $$$$\mathrm{7}^{{b}} \equiv\mathrm{7},\mathrm{9},\mathrm{3}\:{or}\:\mathrm{1}\:\left({mod}\:\mathrm{10}\right);\mathrm{13}^{{d}} \equiv\mathrm{3},\mathrm{9},\mathrm{7}\:{or}\:\mathrm{1}\left({mod}\:\mathrm{10}\right) \\ $$$$\Rightarrow\mathrm{5}^{{a}} +\mathrm{7}^{{b}} +\mathrm{11}^{{c}} +\mathrm{13}^{{d}} \equiv\mathrm{6};\mathrm{2};\mathrm{8};\mathrm{4};\mathrm{0}\:\left({mod}\:\mathrm{10}\right)\Rightarrow\left({c}\right) \\ $$

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