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Math Help - Congruence

  1. #1
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    Congruence

    Demonstrate that 2222^{5555}+5555^{2222} is divisible by 7
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have 2222=7\cdot 317+3, \ 5555=7\cdot 793+4
    Then (7k+3)^{5555}+(7l+4)^{2222}=7p+3^{5555}+4^{2222}=

    =7p+243^{1111}+16^{1111}=7p+259q=7(p+37q)
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  3. #3
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    thank you
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