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    n=28^(112013)-1288.
    (a) Show 7 and 8 divide n. (Done)
    (b) Show 5 and 10 divide n.
    (c) Prove that m|n for all m in {1,2,3,4,5,6,7,8,9,10}.

    I'm completely stuck on parts b and c, though I feel like if I can do one I could work out the other.

    I've been trying to move around the powers and factor out 28 wherever possible but it's been an hour and I'm getting nowhere.

    Also, for part c is there a general way to do it or do I need to prove it separately for all 10??

    Thanks!
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    (Original post by JustJusty)
    n=28^(112013)-1288.
    (a) Show 7 and 8 divide n. (Done)
    (b) Show 5 and 10 divide n.
    (c) Prove that m|n for all m in {1,2,3,4,5,6,7,8,9,10}.

    I'm completely stuck on parts b and c, though I feel like if I can do one I could work out the other.

    I've been trying to move around the powers and factor out 28 wherever possible but it's been an hour and I'm getting nowhere.

    Also, for part c is there a general way to do it or do I need to prove it separately for all 10??

    Thanks!
    For part (b) with the 5 you can say (28^{112013}-1288) \mod 5 = (3)^{112013} \mod 5 - 3\mod 5

    =3 \mod 5 \cdot [3^{112012} \mod 5 -1 \mod 5]

    =3 \mod 5 \cdot [(3^4)^{28003} \mod 5 -1 \mod 5]

    =3 \mod 5 \cdot [(1)^{28003} \mod 5 -1 \mod 5]


    Then do something similar with the 10.

    I think there might be a way to get both cases, 5 and 10, down in one go but I haven't considered it here.
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    (Original post by RDKGames)
    For part (b) with the 5 you can say (28^{112013}-1288) \mod 5 = (3)^{112013} \mod 5 - 3\mod 5

    =3 \mod 5 \cdot [3^{112012} \mod 5 -1 \mod 5]

    =3 \mod 5 \cdot [(3^4)^{28003} \mod 5 -1 \mod 5]

    =3 \mod 5 \cdot [(1)^{28003} \mod 5 -1 \mod 5]


    Then do something similar with the 10.

    I think there might be a way to get both cases, 5 and 10, down in one go but I haven't considered it here.
    Thanks! I don't know why I didn't simplify the 1288 down to 3 since I did it for 28.
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    (Original post by RDKGames)
    I think there might be a way to get both cases, 5 and 10, down in one go but I haven't considered it here.
    Well you've shown it for 5, and it's clearly even,....
 
 
 
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