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Math Help - How do you find the last digit of a number using Fermat's Little Theorem

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    Question How do you find the last digit of a number using Fermat's Little Theorem

    For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
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    Grand Panjandrum
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    Quote Originally Posted by steph3824 View Post
    For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
    I'm not sure why one would need Fermat's little theorem for this since:

    17^3 \equiv 1 \text{ mod } 10

    and:

     17^{3241}=(17^3)^{1080}\times 17

    which immediately should tell you that:

    17^{3241}\equiv 7 \text{ mod } 10

    CB
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    Quote Originally Posted by CaptainBlack View Post
    I'm not sure why one would need Fermat's little theorem for this since:

    17^3 \equiv 1 \text{ mod } 10

    and:

     17^{3241}=(17^3)^{1080}\times 17

    which immediately should tell you that:

    17^{3241}\equiv 7 \text{ mod } 10

    CB
    Incorrect!

    17^4 \equiv 1 (mod 10)

    17^{3241}=(17^4)^{810}\times 17

    So 17^{3241}\equiv 7(mod 10)
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    Grand Panjandrum
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    Quote Originally Posted by alexmahone View Post
    Incorrect!

    17^4 \equiv 1 (mod 10)

    17^{3241}=(17^4)^{810}\times 17

    So 17^{3241}\equiv 7(mod 10)
    Oppsss.. arithmetic error (actually a transcription error with argument completed with the erroeous value)

    CB
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    Quote Originally Posted by steph3824 View Post
    For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
    Here is a solution via Fermat's Last Theorem (although it's not necessarily the simplest):

    By Fermat's Last Theorem, 17^4 \equiv 1 \mod 5,
    so since
    17^{3241} = (17^4)^{810} \cdot 17,
    17^{3241} \equiv 1 \cdot 17 \equiv 2 \mod 5.

    We also have 17 \equiv 1 \mod 2, so 17 ^ {3241} \equiv 1 \mod 2.

    Since 17^{3241} \equiv 2 \mod 5 and 17 ^ {3241} \equiv 1 \mod 2,
    17^{3241} \equiv 7 \mod 10.
    We could use the Chinese Remainder Theorem at the last step, but it seems like overkill here.
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