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Prove-that-1-5555-2222-2222-5555-divisible-by-7-2-3-105-4-105-divisible-by-7-

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Question Number 188247 by cortano12 last updated on 27/Feb/23
Prove that (1) 5555^(2222) +2222^(5555) divisible by 7 (2) 3^(105) +4^(105) divisible by 7
$$\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left(\mathrm{1}\right)\:\mathrm{5555}^{\mathrm{2222}} +\mathrm{2222}^{\mathrm{5555}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{105}} +\mathrm{4}^{\mathrm{105}} \:\mathrm{divisible}\:\mathrm{by}\:\mathrm{7}\: \\ $$
Answered by Rasheed.Sindhi last updated on 27/Feb/23
(1) 5555^(2222) +2222^(5555) ≡0(mod 7) (793×7+4)^(2222) +(317×7+3)^(5555) ≡0(mod 7 4^(2222) +3^(5555) ≡0(mod 7) 4^1 ≡4(mod 7) 3^1 ≡3(mod 7) 4^2 ≡2(mod 7) 3^2 ≡2(mod 7) 4^3 ≡1(mod 7) 3^3 ≡−1(mod 7) 4^6 ≡1(mod 7) 3^6 ≡1(mod 7) 4^(2222) +3^(5555) ≡0(mod 7) 4^(370×6+2) +3^(925×6+5) ≡0(mod 7) 4^2 +3^5 ≡0(mod 7) 16+243≡0(mod 7) 259≡0(mod 7) clearly true (2) 3^(105) +4^(105) ≡0(mod 7) 3^(17×6+3) +4^(17×6+3) ≡0(mod 7) 3^3 +4^3 ≡0(mod 7) 27+64≡0(mod 7) 91≡0(mod 7) 7×13≡(mod 7) clearly true
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{5555}^{\mathrm{2222}} +\mathrm{2222}^{\mathrm{5555}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\left(\mathrm{793}×\mathrm{7}+\mathrm{4}\right)^{\mathrm{2222}} +\left(\mathrm{317}×\mathrm{7}+\mathrm{3}\right)^{\mathrm{5555}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right. \\ $$$$\mathrm{4}^{\mathrm{2222}} +\mathrm{3}^{\mathrm{5555}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{4}^{\mathrm{1}} \equiv\mathrm{4}\left({mod}\:\mathrm{7}\right)\:\:\:\mathrm{3}^{\mathrm{1}} \equiv\mathrm{3}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{4}^{\mathrm{2}} \equiv\mathrm{2}\left({mod}\:\mathrm{7}\right)\:\:\:\:\mathrm{3}^{\mathrm{2}} \equiv\mathrm{2}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{4}^{\mathrm{3}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\:\:\mathrm{3}^{\mathrm{3}} \equiv−\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{4}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right)\:\:\:\mathrm{3}^{\mathrm{6}} \equiv\mathrm{1}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{4}^{\mathrm{2222}} +\mathrm{3}^{\mathrm{5555}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{4}^{\mathrm{370}×\mathrm{6}+\mathrm{2}} +\mathrm{3}^{\mathrm{925}×\mathrm{6}+\mathrm{5}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{5}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{16}+\mathrm{243}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\mathrm{259}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right)\:{clearly}\:{true} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\mathrm{3}^{\mathrm{105}} +\mathrm{4}^{\mathrm{105}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\:\mathrm{3}^{\mathrm{17}×\mathrm{6}+\mathrm{3}} +\mathrm{4}^{\mathrm{17}×\mathrm{6}+\mathrm{3}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\mathrm{3}^{\mathrm{3}} +\mathrm{4}^{\mathrm{3}} \equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\mathrm{27}+\mathrm{64}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\:\mathrm{91}\equiv\mathrm{0}\left({mod}\:\mathrm{7}\right) \\ $$$$\:\:\mathrm{7}×\mathrm{13}\equiv\left({mod}\:\mathrm{7}\right)\:{clearly}\:{true} \\ $$

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