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Math Help - Two quick questions

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    Super Member Aryth's Avatar
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    Two quick questions

    These are stumping me for some odd reason...

    1. Find the remainder of 2009^{2008} upon division by 9.

    2. Determine with explanation, whether there exists an integer n such that n^{10} + 1 is divisible by 151. (Note that 151 is prime)
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  2. #2
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    Hello, Aryth!

    1. Find the remainder of 2009^{2008} upon division by 9.

    Since 2009 \:=\:223(9) + 2, then: . 2009 \div 9 has a remainder of 2.

    Then: . 2009^{2008} \div 9 has a remainder of 2^{2008}


    We find that: . 2^6 \,=\,64 . . . and 64 \div 9 has remainder 1.

    \text{Then: }\;2^{2008} \;=\;2^{334(6) + 4} \;=\;2^{334(6)}\cdot 2^4 \;=\;\underbrace{\left(2^6\right)^{334}}_{\text{re  m. 1}}\cdot16 <br />

    Hence: .  16 \div 9 \quad\to\quad \text{remainder } 7

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    Super Member Aryth's Avatar
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    Would anyone be able to help with the second one?
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    Quote Originally Posted by Aryth View Post

    Would anyone be able to help with the second one?
    well, it's quite easy: since \left(\frac{-1}{151} \right)=-1, the equation x^2 \equiv -1 \mod 151 has no solution and thus, obviously, x^{10} \equiv -1 \mod 151 cannot have any solution either.
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    Quote Originally Posted by NonCommAlg View Post
    well, it's quite easy: since \left(\frac{-1}{151} \right)=-1, the equation x^2 \equiv -1 \mod 151 has no solution and thus, obviously, x^{10} \equiv -1 \mod 151 cannot have any solution either.
    another way: suppose n^{10} \equiv -1 \mod 151 has a solution. then n^{150}=(n^{10})^{15} \equiv -1 \mod 151. but, since \gcd(n,151)=1, by Fermat's little theorem n^{150} \equiv 1 \mod 151. contradiction!
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    Super Member Aryth's Avatar
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    I seriously appreciate the help.
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