9 ^ 100 (mod 11) ?

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  • sepia
    Lv 7
    1 month ago
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  • JOHN
    Lv 7
    1 month ago

    Answer is below

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    • JOHN
      Lv 7
      1 month agoReport

      @Morewood. I know all of that. The idea was to offer something different from the other answers, one that someone who knows no umber theory could understand. I thought I had made that clear in the preamble to my solution.

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  • nbsale
    Lv 6
    1 month ago

    Use Fermat's Little Theorem, and this is trivial:

    If p is prime and (a,p) = 1 then a^(p-1) = 1 mod p

    11 is a prime, and (9,11) = 1 (i.e., they are relatively prime).

    Therefore 9^10 = 1 mod p, and so is 9^100 since 100 is a multiple of 10:

    9^(100) = 9^(10)^10 = 1^10 = 1 mod 11

    Other solutions are OK, but way more work. If this is from a number theory class, you definitely should be using FLT, which is simple and very basic to number theory. You should also know Euler's extension to this, (a,n) = 1 ==> a^phi(n) = 1 mod n.

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  • 1 month ago

    9^100 (mod 11)

    = (9²)^50 (mod 11)

    = 81^50 (mod 11)

    = 4^50 (mod 11)

    = (2²)^50 (mod 11)

    = 2^100 (mod 11)

    = (2^5)^20 (mod 11)

    = (-1)^20 (mod 11)

    = ((-1)²)^10 (mod 11)

    = 1^10 (mod 11)

    = 1 (mod 11)

    Answer:

    1

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  • 1 month ago

    9 mod11 = 9

    9^2 mod11 = 4

    9^3 mod 11 = 3

    9^4 mod11 = 5

    9^5 mod11 = 1

    9^6 mod11 = 9

    Pattern repeats every fifth term.

    100/5 =20 with no remainder

    9^100 mod11 = 1

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