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Math Help - Is this right? (modular arthimetic)

  1. #1
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    Is this right? (modular arthimetic)

    Just want some confirmation on what i've done -number theory's my weakest area by far.

    1. Calculate 7^23 (mod 10)

    MY ANSWER: 7=17 (mod 10)

    7^2 = 289 = 9(mod 10)

    so 7^22 = (7^2)^11 = 9 (mod 10)

    so 7^23 = 7 x 7^22 = 17 x 9^11 = 9(mod 10)
    (note: these are meant to be equivalent rather than equals signs)

    2. Solve 8x = 12 (mod 20)

    MY ANSWER:

    gcd(8,20) = 4
    4 divides 12 so there are 3 solutions.
    A solution is x = 14 (by observation).
    Solutions differ by 20/4 = 5.
    So 9 and 4 are other solutions as general solutuons given by x= u - (n/d)t
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  2. #2
    Senior Member Dinkydoe's Avatar
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    First:

    Observe that:

    7^2= 49 \equiv 9 mod 10
    7^4\equiv 9^2\equiv 1 mod 10

    so we observe that: 7^{a+4k}\equiv 7^amod 10

    Thus 7^{23}\equiv 7^3\cdot 7^{20}\equiv 7^3\equiv 7\cdot 9\equiv 3 mod 10

    So you answer is incorrect. And the steps you take don't allways make sense.

    2.
    8x\equiv 12 mod 20
    8x-12\equiv 8x+8\equiv 0 mod 20
    8(x+1)\equiv 0 mod 20


    Observe that 8(x+1)\equiv 0 mod 20 \Leftrightarrow 5|(x+1)


    Thus:
    x\equiv 4 mod 20
    x\equiv  9 mod 20
    x\equiv 14 mod 20
    x\equiv 19 mod 20
    Last edited by Dinkydoe; January 15th 2010 at 10:59 AM.
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  3. #3
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    <br />
7^{23} = 7^{2^4+2^2+2^1+2^0} = 7^{2^4} \cdot 7^{2^2} \cdot 7^{2^1} \cdot 7^{2^0} =  <br />
7^{(2\cdot2\cdot2\cdot2)} \cdot 7^{(2\cdot2)} \cdot 7^2 \cdot 7 =<br />
(((7^2)^2)^2)^2 \cdot (7^2)^2 \cdot 7^2 \cdot 7<br />

     7^2 = 49 \equiv  = 9\pmod{10}

    Substitute: above into equation <br />
7^{23} \equiv (((7^2)^2)^2)^2 \cdot (7^2)^2 \cdot 7^2 \cdot 7 \pmod{10} gives us

    <br />
7^{23} \equiv ((9^2)^2)^2 \cdot 9^2 \cdot 9 \cdot 7 \pmod{10}<br />

    Do this process again for  9^2 = 81 \equiv 1 \pmod{10} . We get:

    <br />
7^{23} \equiv ((1)^2)^2 \cdot 1 \cdot 9 \cdot 7 \pmod{10}<br />

    So finally,

    <br />
7^{23} \equiv 1 \cdot 1 \cdot 9 \cdot 7 \pmod{10} \implies<br />
7^{23} \equiv 63 \pmod{10} \implies <br />
7^{23} \equiv 3 \pmod{10}<br />

    -Andy
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  4. #4
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    Quote Originally Posted by Dinkydoe View Post
    First:

    Observe that:

    7^2= 49 \equiv 9 mod 10
    7^4\equiv 9^2\equiv 1 mod 10

    so we observe that: 7^{a+4k}\equiv 7^amod 10

    Thus 7^{23}\equiv 7^3\cdot 7^{20}\equiv 7^3\equiv 7\cdot 9\equiv 63 mod 10 = 3 mod 10
    EDIT: You corrected your typo. So this post is irrelevant.
    -Andy
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Yep: True,

    My answer was correct: It was a typo, as you can see I allready corrected it:

    The deduction to the last step was correct: \equiv 9\cdot 7 = 63 \equiv 3 mod 10
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  6. #6
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    Quote Originally Posted by Dinkydoe View Post
    Yep: True,

    My answer was correct: It was a typo, as you can see I allready corrected it:

    The deduction to the last step was correct: \equiv 9\cdot 7 = 63 \equiv 3 mod 10
    My fault. The huge red font was not what I wanted. It looks super-obnoxious. I'm going to scale it down.

    -Andy
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